MIT grad shows how to use the Law of Sines to solve a triangle for missing sides and angles. To skip ahead: 1) for the AAS or ASA case, when you're given TWO ANGLES AND A SIDE, skip to 0:17. 2) For how to know WHEN to use the Law of Sines in trigonometry, skip to 5:55. 3) For when the SSA ambiguous case comes up, and there may be one solution, two solutions, or no solution, skip to 6:32. Nancy explains the trig steps to find the missing sides of a triangle using the sine function in this basic introduction to the Law of Sines formula used in trigonometry, geometry, algebra 2, and precalculus. This trig law is also known as the Sine Rule. It is helpful for solving OBLIQUE triangles with trigonometric functions. For help with and examples of solving a RIGHT triangle using trig functions, jump to https://youtu.be/a5WQlcFTXyk. For an intro to TRIGONOMETRY BASICS (sin, cos, tan, csc, sec, and cot) and SOH-CAH-TOA, jump to: https://youtu.be/bSM7RNSbWhM.

If you need to "solve the triangle" in a trigonometry problem, it just means to find all the missing angles and sides of the triangle.

WHAT is the LAW of SINES? Say that you're given an oblique triangle, meaning one that is not a right angle triangle, and you need to solve it. If you know two angles and at least one of the sides, you can solve with the Law of Sines trigonometry formula. The Law of Sines says that the sine of one angle over the side opposite that angle, equals the sine of the next angle over its opposite side, which equals sine of the last angle over its opposite length. These three ratios equal each other, and you can use these proportions to solve for unknowns.

As an example, say that you're given that angle B is 34 degrees, angle C is 110 degrees, and side b is 14. We know two angles and one side. What's missing, or what we're looking for, is side a, side c, and the measure of angle A. That missing third angle, angle A, we can find very quickly without using any special new law or trig identity, since the sum of the degrees in a triangle is always 180 degrees. Subtracting the two known angles (34 and 110 degrees) from the total 180 degrees gives that the measure of angle A is 36 degrees.

HOW to use the Law of Sines: to find the missing two side lengths, we do need the Law of Sines theorem. First write all the ratios from the Law of Sines: the sine of each angle, over the length opposite. We write the sine of 36 degrees over the opposite side, side a. This equals sine of the next angle, 34 degrees, over its opposite side 14. And finally, that equals sine of the last angle, 110 degrees, over the opposite length, the unknown c.

If we look at just the leftmost two ratios, we see that we can use that equation to solve for a. So we separate that part out to solve: sin(36)/a = sin(34)/14. An easy way to solve a proportion like this is to cross-multiply. This gives us 14sin(36) = a*sin(34). Now it's more clear how to solve for side a, by just dividing out sin(34) from both sides. So we get that a is equal to 14*sin(36)/sin(34). To get a number value that's practical as the length of a triangle side, you can use your calculator to get a decimal number. CAUTION: since we were given angles in degrees, make sure your calculator is in degree mode, not radian mode, so that you don't get the wrong answer. The side length a is then approximately 14.72.

Now there's just one more unknown to find, the side c. To find the remaining missing side of a triangle, you can use a different pair of ratios to solve, the equation with the rightmost two ratios: sin(34)/14 = sin(110)/c. In general, to solve for one of the unknowns, use the ratio that has what you want to find in it, as the only unknown, and set it equal to a ratio where you know everything already, and you will get the answer. We cross-multiply to get c*sin(34) = 14sin(110). When we get c alone and use a calculator, we find that side length c is approx. 23.53. Now we've completely solved the triangle for all the missing sides and angles.

How do you know WHEN to use the LAW of SINES? If you're given two angles and a side (AAS or ASA cases), or two sides and an angle opposite one of those sides (SSA), you can use the Law of Sines property. If you have two angles and one side (AAS/ASA), you can use the Law of Sines just as we did to find the missing sides. WARNING: for the SSA case, when you have two sides and an angle that's opposite one of them, you can also use the Law of Sines to solve, but instead of having one solution, there may be no solution or two solutions. The SSA case is also called the "ambiguous case".

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

If you need to "solve the triangle" in a trigonometry problem, it just means to find all the missing angles and sides of the triangle.

WHAT is the LAW of SINES? Say that you're given an oblique triangle, meaning one that is not a right angle triangle, and you need to solve it. If you know two angles and at least one of the sides, you can solve with the Law of Sines trigonometry formula. The Law of Sines says that the sine of one angle over the side opposite that angle, equals the sine of the next angle over its opposite side, which equals sine of the last angle over its opposite length. These three ratios equal each other, and you can use these proportions to solve for unknowns.

As an example, say that you're given that angle B is 34 degrees, angle C is 110 degrees, and side b is 14. We know two angles and one side. What's missing, or what we're looking for, is side a, side c, and the measure of angle A. That missing third angle, angle A, we can find very quickly without using any special new law or trig identity, since the sum of the degrees in a triangle is always 180 degrees. Subtracting the two known angles (34 and 110 degrees) from the total 180 degrees gives that the measure of angle A is 36 degrees.

HOW to use the Law of Sines: to find the missing two side lengths, we do need the Law of Sines theorem. First write all the ratios from the Law of Sines: the sine of each angle, over the length opposite. We write the sine of 36 degrees over the opposite side, side a. This equals sine of the next angle, 34 degrees, over its opposite side 14. And finally, that equals sine of the last angle, 110 degrees, over the opposite length, the unknown c.

If we look at just the leftmost two ratios, we see that we can use that equation to solve for a. So we separate that part out to solve: sin(36)/a = sin(34)/14. An easy way to solve a proportion like this is to cross-multiply. This gives us 14sin(36) = a*sin(34). Now it's more clear how to solve for side a, by just dividing out sin(34) from both sides. So we get that a is equal to 14*sin(36)/sin(34). To get a number value that's practical as the length of a triangle side, you can use your calculator to get a decimal number. CAUTION: since we were given angles in degrees, make sure your calculator is in degree mode, not radian mode, so that you don't get the wrong answer. The side length a is then approximately 14.72.

Now there's just one more unknown to find, the side c. To find the remaining missing side of a triangle, you can use a different pair of ratios to solve, the equation with the rightmost two ratios: sin(34)/14 = sin(110)/c. In general, to solve for one of the unknowns, use the ratio that has what you want to find in it, as the only unknown, and set it equal to a ratio where you know everything already, and you will get the answer. We cross-multiply to get c*sin(34) = 14sin(110). When we get c alone and use a calculator, we find that side length c is approx. 23.53. Now we've completely solved the triangle for all the missing sides and angles.

How do you know WHEN to use the LAW of SINES? If you're given two angles and a side (AAS or ASA cases), or two sides and an angle opposite one of those sides (SSA), you can use the Law of Sines property. If you have two angles and one side (AAS/ASA), you can use the Law of Sines just as we did to find the missing sides. WARNING: for the SSA case, when you have two sides and an angle that's opposite one of them, you can also use the Law of Sines to solve, but instead of having one solution, there may be no solution or two solutions. The SSA case is also called the "ambiguous case".

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

MIT grad shows how to use the chain rule for EXPONENTIAL, LOG, and ROOT forms and how to use the chain rule with the PRODUCT RULE to find the derivative. To skip ahead: 4) For an example with an EXPONENTIAL function that needs the chain rule to take the derivative, skip to time 0:32. 5) For an example with a natural LOG, skip to 2:34. 6) For an example with a SQUARE ROOT, skip to 4:41. 7) For how to use the CHAIN RULE with the PRODUCT RULE, and how to know which to use first, skip to 7:17. For an introduction of HOW and WHEN to use the chain rule, jump to my FIRST chain rule video: https://youtu.be/H-ybCx8gt-8.

The CHAIN RULE is one of the derivative rules. You need it to take the derivative when you have a function inside a function, or a "composite function". This video picks up at the end of the first chain rule video ("Chain Rule...How? When?" at https://youtu.be/H-ybCx8gt-8) and shows more examples of using the chain rule, as well as how to combine the chain rule with the product rule.

For an intro of how to take derivatives, jump to: https://youtu.be/QqF3i1pnyzU

4) EXPONENTIAL: for the example y = e^(3x), first take the derivative of the outside function (the exponential function, e). The derivative of the exponential function is just the exponential function itself. When writing the outside derivative, leave the inside function (3x) the same, so write e^(3x). Then by the chain rule, you multiply by the derivative of just the inner function (3x). Since the inside derivative is 3, the answer for the full derivative dy/dx is 3e^(3x).

5) NATURAL LOG example: To find the derivative of y = ln(5x), use the same steps as above to first take the outside derivative and then multiply by the inside derivative. 5x is inside the natural log. The outer function here is the natural log, ln. To take the derivative of the natural log of something, we write 1 over that something, so 1 over 5x. We leave the inside function alone here when writing the outside derivative, so we write 1/(5x). Remember that we're not done yet. We also need to multiply by the derivative of the inside function. The derivative of 5x is just 5, so our final simplified answer for dy/dx is 1/x.

6) SQUARE ROOT example: For this example, y = sqrt(x^2 + 1), the x^2 + 1 is under a square root. The square root is the outside, outer function. When you have a root of any kind, before you take the derivative, it's easiest to rewrite it in the form of a fractional power, so that you can use the POWER RULE when taking its derivative. The outside function here is then the 1/2 power, and the inside function is x^2 + 1. To take the derivative of the outside one-half power, we use the power rule to bring down the 1/2 power to the front and reduce the old 1/2 power by 1. We then multiply by the inside derivative, and after simplifying get a derivative of y' = x / sqrt(x^2 + 1).

Here I used Lagrange notation (y') for the derivative instead of the Liebniz notation from before (dy/dx). To learn the POWER RULE, jump to my video at: https://youtu.be/TgIl15Nlg_U

7) CHAIN RULE WITH THE PRODUCT RULE: Sometimes you might need to use the chain rule combined with the product rule. How do you know whether to apply the chain rule first, or the product rule first? In this example, y = x^3 (2x - 5)^4, although the second factor here will need the chain rule for its derivative, overall in the biggest view of this function, we have a product of two factors. So we will use the product rule first here on the two factors, and then use the chain rule inside the product rule. However, for an example like ln[x^3 (2x-5)^4)], you would need to apply the chain rule first on the ln form, and then since your inside function is a product, you would need the product rule after that. Basically, the first thing you'll do is whatever you need to use to handle the outermost form of your function.

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

The CHAIN RULE is one of the derivative rules. You need it to take the derivative when you have a function inside a function, or a "composite function". This video picks up at the end of the first chain rule video ("Chain Rule...How? When?" at https://youtu.be/H-ybCx8gt-8) and shows more examples of using the chain rule, as well as how to combine the chain rule with the product rule.

For an intro of how to take derivatives, jump to: https://youtu.be/QqF3i1pnyzU

4) EXPONENTIAL: for the example y = e^(3x), first take the derivative of the outside function (the exponential function, e). The derivative of the exponential function is just the exponential function itself. When writing the outside derivative, leave the inside function (3x) the same, so write e^(3x). Then by the chain rule, you multiply by the derivative of just the inner function (3x). Since the inside derivative is 3, the answer for the full derivative dy/dx is 3e^(3x).

5) NATURAL LOG example: To find the derivative of y = ln(5x), use the same steps as above to first take the outside derivative and then multiply by the inside derivative. 5x is inside the natural log. The outer function here is the natural log, ln. To take the derivative of the natural log of something, we write 1 over that something, so 1 over 5x. We leave the inside function alone here when writing the outside derivative, so we write 1/(5x). Remember that we're not done yet. We also need to multiply by the derivative of the inside function. The derivative of 5x is just 5, so our final simplified answer for dy/dx is 1/x.

6) SQUARE ROOT example: For this example, y = sqrt(x^2 + 1), the x^2 + 1 is under a square root. The square root is the outside, outer function. When you have a root of any kind, before you take the derivative, it's easiest to rewrite it in the form of a fractional power, so that you can use the POWER RULE when taking its derivative. The outside function here is then the 1/2 power, and the inside function is x^2 + 1. To take the derivative of the outside one-half power, we use the power rule to bring down the 1/2 power to the front and reduce the old 1/2 power by 1. We then multiply by the inside derivative, and after simplifying get a derivative of y' = x / sqrt(x^2 + 1).

Here I used Lagrange notation (y') for the derivative instead of the Liebniz notation from before (dy/dx). To learn the POWER RULE, jump to my video at: https://youtu.be/TgIl15Nlg_U

7) CHAIN RULE WITH THE PRODUCT RULE: Sometimes you might need to use the chain rule combined with the product rule. How do you know whether to apply the chain rule first, or the product rule first? In this example, y = x^3 (2x - 5)^4, although the second factor here will need the chain rule for its derivative, overall in the biggest view of this function, we have a product of two factors. So we will use the product rule first here on the two factors, and then use the chain rule inside the product rule. However, for an example like ln[x^3 (2x-5)^4)], you would need to apply the chain rule first on the ln form, and then since your inside function is a product, you would need the product rule after that. Basically, the first thing you'll do is whatever you need to use to handle the outermost form of your function.

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

Nancy's not perfect! A lot of you have asked if she's ever failed a test. This one's for you. ♡ For the shock, skip to time 0:34! For the anger, skip to 1:49. For when you're feeing stuck, skip to 2:11. For the important thing to remember if you've just failed a test, skip to 2:58.

If you've failed a test, you're not alone. You're in good company, because we've all failed at things in our lives. It doesn't mean you're not capable of the math. Failing a math test does not define you or your whole math experience.

Have you ever failed a test? What advice would you give to someone?

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

Director: Kristopher Knight

https://kristopherknight.com

If you've failed a test, you're not alone. You're in good company, because we've all failed at things in our lives. It doesn't mean you're not capable of the math. Failing a math test does not define you or your whole math experience.

Have you ever failed a test? What advice would you give to someone?

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

Director: Kristopher Knight

https://kristopherknight.com

MIT grad shows how to find the derivative using the Power Rule, one of the derivative rules in calculus. It is a shortcut for taking derivatives of polynomial functions with powers of x. To skip ahead: 1) For HOW and WHEN to use the power rule, skip to time 0:22. 2) For how to use the power rule when you have a FRACTIONAL or NEGATIVE POWER, skip to 5:22. For my video on the other differentiation rules, PRODUCT RULE and QUOTIENT RULE, skip to https://youtu.be/QqF3i1pnyzU?t=456

HOW and WHEN to use the POWER RULE: If the given equation is a polynomial, or just a power of x, then you can use the Power Rule. For a term that's just a power of x, such as x^4, you can get the derivative by bringing down the power to the front of the term as a coefficient and decreasing the x power by 1. For example, for x^4, the derivative is 4x^3. If you have many terms added or subtracted together, and if they are powers of x, you can use the Power Rule on each term (by the Sum and Difference Rules).

NOTE: The derivative of a constant, just a number, is always 0 (that is the Constant Rule). Also, if you have a term that is a constant multiplied in the front of the term, like 2x^3, you can keep the constant and differentiate the rest of the term (Constant Multiple Rule). In this example, you keep the 2 and take the derivative of x^3, which is 3x^2, so the derivative of the term 2x^3 is 2*3x^2, or 6x^2.

ANOTHER NOTE: You can use the same power rule method for fractional or negative powers, but be careful... for negative powers, it works as long as x is not 0, and for fractional/rational powers, if the power is less than 1, your derivative won't be defined at x = 0.

The derivative is a function that gives you the instantaneous rate of change at each point of another function. You can calculate the derivative with the definition of the derivative (using the limit, see https://youtu.be/-ktrtzYVk_I?t=628), but the fastest way to find the derivative is with shortcuts such as the Power Rule, Product Rule, and Quotient Rule.

HOW and WHEN to use the POWER RULE: If the given equation is a polynomial, or just a power of x, then you can use the Power Rule. For a term that's just a power of x, such as x^4, you can get the derivative by bringing down the power to the front of the term as a coefficient and decreasing the x power by 1. For example, for x^4, the derivative is 4x^3. If you have many terms added or subtracted together, and if they are powers of x, you can use the Power Rule on each term (by the Sum and Difference Rules).

NOTE: The derivative of a constant, just a number, is always 0 (that is the Constant Rule). Also, if you have a term that is a constant multiplied in the front of the term, like 2x^3, you can keep the constant and differentiate the rest of the term (Constant Multiple Rule). In this example, you keep the 2 and take the derivative of x^3, which is 3x^2, so the derivative of the term 2x^3 is 2*3x^2, or 6x^2.

ANOTHER NOTE: You can use the same power rule method for fractional or negative powers, but be careful... for negative powers, it works as long as x is not 0, and for fractional/rational powers, if the power is less than 1, your derivative won't be defined at x = 0.

The derivative is a function that gives you the instantaneous rate of change at each point of another function. You can calculate the derivative with the definition of the derivative (using the limit, see https://youtu.be/-ktrtzYVk_I?t=628), but the fastest way to find the derivative is with shortcuts such as the Power Rule, Product Rule, and Quotient Rule.

MIT grad shows how to solve for the sides and angles of a right triangle using trig functions and how to find the missing sides of a right triangle with trigonometry basics. To skip ahead: 1) For HOW to CHOOSE A TRIG FUNCTION to solve for a side you don't know, skip to time 2:10. 2) For how to SOLVE for the LAST SIDE with TRIG, skip to time 6:18. 3) For how to SOLVE for the LAST SIDE with the PYTHAGOREAN THEOREM, skip to time 8:41. 4) For how to SOLVE FOR THE LAST ANGLE, skip to time 10:38. For my video that introduces the BASIC TRIG functions (sin, cos, tan, csc, sec, and cot) and SOH-CAH-TOA, jump to: https://youtu.be/bSM7RNSbWhM

HOW TO SOLVE A RIGHT TRIANGLE: If you have to "solve a right triangle," it just means to find all the missing angles and sides of a triangle. When you have a right triangle and you already know at least one of the angles (other than the right angle) and at least one of the sides, you can solve for the other sides using a trig function: sine, cosine, or tangent. When solving right triangles, you can use the memory trick sohcahtoa (SOH CAH TOA) to remember the trig ratios for sin, cos, and tan. Pick the trig function that includes both the side you already know AND the side you're looking for. You can use algebra to solve for the length of the unknown side, to find the missing side of a triangle. Once you have that side, you have two options for finding the last unknown side. You can either: 1) use another trig function, or 2) use the Pythagorean Theorem formula. Then you will have all the sides. Make sure you have found all the angles inside the triangle so that you've really "solved" the whole triangle. If you know two angles already (one of them is the right angle, 90 degrees), you can find the last unknown angles really quickly. Since all three angles have to add up to 180 degrees, you can subtract the two angles from 180 to find the unknown angle in degrees.

HOW TO SOLVE A RIGHT TRIANGLE: If you have to "solve a right triangle," it just means to find all the missing angles and sides of a triangle. When you have a right triangle and you already know at least one of the angles (other than the right angle) and at least one of the sides, you can solve for the other sides using a trig function: sine, cosine, or tangent. When solving right triangles, you can use the memory trick sohcahtoa (SOH CAH TOA) to remember the trig ratios for sin, cos, and tan. Pick the trig function that includes both the side you already know AND the side you're looking for. You can use algebra to solve for the length of the unknown side, to find the missing side of a triangle. Once you have that side, you have two options for finding the last unknown side. You can either: 1) use another trig function, or 2) use the Pythagorean Theorem formula. Then you will have all the sides. Make sure you have found all the angles inside the triangle so that you've really "solved" the whole triangle. If you know two angles already (one of them is the right angle, 90 degrees), you can find the last unknown angles really quickly. Since all three angles have to add up to 180 degrees, you can subtract the two angles from 180 to find the unknown angle in degrees.

Nancy, formerly of mathbff, shows how to simplify a rational expression. To skip ahead: 1) for how to simplify an expression with DIFFERENCE of SQUARES factoring in it (x^2 - 49), skip to time 1:01. 2) for how to write the CONDITION or RESTRICTION, skip to time 3:52. 3) for how to simplify an expression with QUADRATIC(s) in it (like the x^2 - 3x + 2 trinomial), skip to time 6:12. 4) for an example with LARGER COEFFICIENTS and a NEGATIVE x^2 term (like -20x^2 + 14x + 12), skip to time 9:57.

For how to FACTOR QUADRATICS, jump to https://youtu.be/YtN9_tCaRQc

For how to FIND the DOMAIN of a function, jump to https://youtu.be/GQGFMUfr10M

For more factoring examples in my video on how to solve quadratic equations by FACTORING, jump to: https://youtu.be/Z5MnP9da4EM

There are three steps to simplify algebraic rational expressions:

1) Factor the numerator and denominator as much as possible.

2) Cancel factors that are the same on top and bottom.

3) Write a restriction ("condition") for any x values that used to make the denominator zero, before you cancelled factors. For instance: "x cannot equal -1".

1) FACTOR as much as you can: Factor the top and bottom as much as possible. In both the numerator and denominator, always check first if there's a number or a variable (x) that is in all the terms that you can pull out front. If you have something that looks like "x^2 minus a perfect square number", like x^2 - 9, you can factor using the difference of squares formula, into (x + 3)(x - 3). If you have a difference of cubes (x^3 - 8) or sum of cubes (x^3 + 27), you can factor using the sum/difference of cubes formulas. If you have a quadratic expression on top or bottom that has three terms ("trinomial"), you can factor it using trial and error, or the "magic x" trick for factoring. To learn how to factor quadratics, check out my video on factoring.

2) CANCEL: Cross out any factors that are the same on top and bottom, and write what's left behind. When you've cancelled everything you can, you now have your simplified rational expression, or expression "reduced to lowest terms" or in "simplest form".

3) CONDITION: Is that all? To be technically correct, you should also write a note next to the simplified rational expression that states the restriction. You can check with your teacher to see if you’re required to write it for your class. But either way, the restriction is there, and the final answer isn’t really complete without this “condition”. All you need to do is: look at any factor(s) you cancelled in the denominator (ex: x-7), see what x value used to make that factor equal to zero (7), and next to your answer write “x not equal to” that number (x not equal to 7).

We put the restriction, or condition, there at the end, because if we didn’t, the restricted x value would appear to be a valid x value for the expression, when it’s not. When we simplified the expression by cancelling the bottom factor, we got rid of the undefined, division-by-zero problem for that x-value, but in reality that x-value is still not allowed. So to be totally correct, write the note that states this restriction next to your answer.

For how to find the HORIZONTAL ASYMPTOTE of a rational function, jump to https://youtu.be/qJrrZQgSkO8

For how to FACTOR QUADRATICS, jump to https://youtu.be/YtN9_tCaRQc

For how to FIND the DOMAIN of a function, jump to https://youtu.be/GQGFMUfr10M

For more factoring examples in my video on how to solve quadratic equations by FACTORING, jump to: https://youtu.be/Z5MnP9da4EM

There are three steps to simplify algebraic rational expressions:

1) Factor the numerator and denominator as much as possible.

2) Cancel factors that are the same on top and bottom.

3) Write a restriction ("condition") for any x values that used to make the denominator zero, before you cancelled factors. For instance: "x cannot equal -1".

1) FACTOR as much as you can: Factor the top and bottom as much as possible. In both the numerator and denominator, always check first if there's a number or a variable (x) that is in all the terms that you can pull out front. If you have something that looks like "x^2 minus a perfect square number", like x^2 - 9, you can factor using the difference of squares formula, into (x + 3)(x - 3). If you have a difference of cubes (x^3 - 8) or sum of cubes (x^3 + 27), you can factor using the sum/difference of cubes formulas. If you have a quadratic expression on top or bottom that has three terms ("trinomial"), you can factor it using trial and error, or the "magic x" trick for factoring. To learn how to factor quadratics, check out my video on factoring.

2) CANCEL: Cross out any factors that are the same on top and bottom, and write what's left behind. When you've cancelled everything you can, you now have your simplified rational expression, or expression "reduced to lowest terms" or in "simplest form".

3) CONDITION: Is that all? To be technically correct, you should also write a note next to the simplified rational expression that states the restriction. You can check with your teacher to see if you’re required to write it for your class. But either way, the restriction is there, and the final answer isn’t really complete without this “condition”. All you need to do is: look at any factor(s) you cancelled in the denominator (ex: x-7), see what x value used to make that factor equal to zero (7), and next to your answer write “x not equal to” that number (x not equal to 7).

We put the restriction, or condition, there at the end, because if we didn’t, the restricted x value would appear to be a valid x value for the expression, when it’s not. When we simplified the expression by cancelling the bottom factor, we got rid of the undefined, division-by-zero problem for that x-value, but in reality that x-value is still not allowed. So to be totally correct, write the note that states this restriction next to your answer.

For how to find the HORIZONTAL ASYMPTOTE of a rational function, jump to https://youtu.be/qJrrZQgSkO8

Q&A with Nancy! Nancy, formerly of MathBFF, answers your questions on her new math channel, NancyPi.

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

Director: Kristopher Knight

https://kristopherknight.com

Editor: Miriam Nielsen of zentouro

https://www.youtube.com/zentouro

Director: Kristopher Knight

https://kristopherknight.com

Nancy, formerly of mathbff, shows how to do a binomial expansion with the Binomial Theorem and/or Pascal's Triangle. To skip ahead: 1) for HOW TO EXPAND a BINOMIAL raised to a power, like (x + 3)^5, skip to time 0:57; 2) for how to find the BINOMIAL COEFFICIENTS with the FACTORIAL/COMBINATION formula, skip to time 03:29; 3) to use PASCAL'S TRIANGLE to find binomial coefficients for expansion, skip to time 09:32; 4) for how to write the expansion for a SUBTRACTION/DIFFERENCE binomial raised to a power, skip to time 13:11.

1) HOW TO START A BINOMIAL EXPANSION: If your binomial is something like (x + 3) raised to a power like (x + 3)^5, you have two parts of your binomial: x and 3. You're going to take each of those and raise them to different powers in each term of the expansion. You can start your expansion just by writing these powers out.. For the FIRST PART of your binomial, x, it will start with a power of 5 (your power number) in the first term of the expansion (x^5), and then in each term after the power will go down by 1 (x^4, x^3, etc. all the way down to a zero power, x^0). Then you take the SECOND PART of the binomial, 3, and multiply each term by a power of 3. The 3 factor will start with a power of 0 in the first term, so 3^0, and then will increase by 1 power in each term after.

2) HOW TO FIND THE COEFFICIENTS (with the FACTORIAL or COMBINATION method): Then there's one more number to find, a number that gets multiplied in front of each term, or a binomial coefficient. There are two ways to find the coefficients (for the Pascal's Triangle way, see below). To find the coefficients using the factorial, combination ("n choose k") formula of n!/(k!(n-k)!, each term has a coefficient number you find using an n value equal to the power number, 5, and a k value that runs from 0 for the first term up through 5 for the last term. This number gets multiplied by the other factors in each term, and then simplify for your final expansion.

3) HOW TO FIND THE COEFFICIENTS (with PASCAL'S TRIANGLE): You can instead use Pascal's Triangle to find the binomial coefficients. Whichever row of Pascal's Triangle has your power number in it, as the second number, is the row that gives you all the coefficient numbers you'll need for your expansion. Each coefficient is multiplied with its term, and then you can simplify the expansion.

4) WHAT IF THE BINOMIAL HAS SUBTRACTION? It's very similar. It's easier to think of the subtracted term as "adding a negative number", and then all of the negative number (in parentheses) will be raised to the power in each term of the expansion. Note: if the binomial have something like 2x - y instead of x - y, make sure that all of (2x) is raised to each power.

1) HOW TO START A BINOMIAL EXPANSION: If your binomial is something like (x + 3) raised to a power like (x + 3)^5, you have two parts of your binomial: x and 3. You're going to take each of those and raise them to different powers in each term of the expansion. You can start your expansion just by writing these powers out.. For the FIRST PART of your binomial, x, it will start with a power of 5 (your power number) in the first term of the expansion (x^5), and then in each term after the power will go down by 1 (x^4, x^3, etc. all the way down to a zero power, x^0). Then you take the SECOND PART of the binomial, 3, and multiply each term by a power of 3. The 3 factor will start with a power of 0 in the first term, so 3^0, and then will increase by 1 power in each term after.

2) HOW TO FIND THE COEFFICIENTS (with the FACTORIAL or COMBINATION method): Then there's one more number to find, a number that gets multiplied in front of each term, or a binomial coefficient. There are two ways to find the coefficients (for the Pascal's Triangle way, see below). To find the coefficients using the factorial, combination ("n choose k") formula of n!/(k!(n-k)!, each term has a coefficient number you find using an n value equal to the power number, 5, and a k value that runs from 0 for the first term up through 5 for the last term. This number gets multiplied by the other factors in each term, and then simplify for your final expansion.

3) HOW TO FIND THE COEFFICIENTS (with PASCAL'S TRIANGLE): You can instead use Pascal's Triangle to find the binomial coefficients. Whichever row of Pascal's Triangle has your power number in it, as the second number, is the row that gives you all the coefficient numbers you'll need for your expansion. Each coefficient is multiplied with its term, and then you can simplify the expansion.

4) WHAT IF THE BINOMIAL HAS SUBTRACTION? It's very similar. It's easier to think of the subtracted term as "adding a negative number", and then all of the negative number (in parentheses) will be raised to the power in each term of the expansion. Note: if the binomial have something like 2x - y instead of x - y, make sure that all of (2x) is raised to each power.

MIT grad shows how to integrate using trigonometric substitution. To skip ahead: 1) For HOW TO KNOW WHICH trig substitution to use (sin, tan, or sec), skip to 0:18. 2) For how to do a SIN SUB and WHEN TO TRY TRIG SUB, skip to 1:29. 3) For a TAN SUB with the radical in the NUMERATOR, skip to 14:41. 4) For a SEC SUB with a RATIONAL/FRACTIONAL POWER, skip to 22:36. 5) For miscellaneous types, skip to time 24:54. Trig substitution integration is a calculus technique for integrals.

For my video on how to do BASIC INTEGRATION, jump to: https://youtu.be/e1nxhJQyLYI

For my video on how to integrate using U-SUBSTITUTION, jump to: https://youtu.be/8B31SAk1nD8

For my video on how to do INTEGRATION BY PARTS, jump to: https://youtu.be/KKg88oSUv0o

HOW TO KNOW WHICH trig substitution to use: If the x-expression under the radical has the form of "a number minus an x^2 term", then you'll use a SINE substitution. If it has the reversed form of "an x^2 term minus a number", then you'll use a SECANT substitution. And finally, if the expression under the root is addition (the form of "an x^2 term plus a number"), you'll use a TANGENT substitution for the problem.

WHEN to use TRIG SUB: If you have an integral with a radical expression in it, and you know you cannot integrate using: a Table of Integrals integration rule, or using U-Substitution, or using the Power Rule, or using other integration techniques like Integration by Parts or Partial Fractions, then it is a good idea to try trig substitution.

HOW TO DO TRIG SUB. Here are the main steps of trig sub problems: 1) Decide which trig substitution you need [x = asin(theta), x = atan(theta), or x = asec(theta)], filling in the value for "a". 2) Simplify the radical expression by plugging in your substitution for x in the radical [For example, for a sin sub, plug in for x in sqrt(a^2-x^2)]. 3) Find dx from your x substitution. 4) Substitute into the original integral, replacing the radical, the dx, and any other x factors, with the theta expressions you found. 5) Simplify and integrate, using whatever technique you have to use. 6) After you integrate, it's still in terms of theta. So the last step is to get it all back in terms of x. Draw a right triangle off to the side, label the sides with what you know from your trig substitution expression, find the third side, and use the sides of the triangle to get an expression for whatever trig function you need in your final answer.

For my video on how to do BASIC INTEGRATION, jump to: https://youtu.be/e1nxhJQyLYI

For my video on how to integrate using U-SUBSTITUTION, jump to: https://youtu.be/8B31SAk1nD8

For my video on how to do INTEGRATION BY PARTS, jump to: https://youtu.be/KKg88oSUv0o

HOW TO KNOW WHICH trig substitution to use: If the x-expression under the radical has the form of "a number minus an x^2 term", then you'll use a SINE substitution. If it has the reversed form of "an x^2 term minus a number", then you'll use a SECANT substitution. And finally, if the expression under the root is addition (the form of "an x^2 term plus a number"), you'll use a TANGENT substitution for the problem.

WHEN to use TRIG SUB: If you have an integral with a radical expression in it, and you know you cannot integrate using: a Table of Integrals integration rule, or using U-Substitution, or using the Power Rule, or using other integration techniques like Integration by Parts or Partial Fractions, then it is a good idea to try trig substitution.

HOW TO DO TRIG SUB. Here are the main steps of trig sub problems: 1) Decide which trig substitution you need [x = asin(theta), x = atan(theta), or x = asec(theta)], filling in the value for "a". 2) Simplify the radical expression by plugging in your substitution for x in the radical [For example, for a sin sub, plug in for x in sqrt(a^2-x^2)]. 3) Find dx from your x substitution. 4) Substitute into the original integral, replacing the radical, the dx, and any other x factors, with the theta expressions you found. 5) Simplify and integrate, using whatever technique you have to use. 6) After you integrate, it's still in terms of theta. So the last step is to get it all back in terms of x. Draw a right triangle off to the side, label the sides with what you know from your trig substitution expression, find the third side, and use the sides of the triangle to get an expression for whatever trig function you need in your final answer.

MIT grad shows how to find antiderivatives, or indefinite integrals, using basic integration rules. To skip ahead: 1) For how to integrate a polynomial with the POWER RULE, skip to 1:35. 2) For how to integrate NEGATIVE POWERS of x, FRACTIONAL POWERS of x, and RADICALS/ROOTS, skip to 6:12. 3) For how to integrate x^(-1), or 1/x, with the LOG RULE, skip to 7:23. 4) For examples where you use more algebra to rewrite before integrating, to SEPARATE the numerator of a FRACTION, or EXPAND a PRODUCT in order to use the Power Rule, skip to 10:00. 5) For basic TRIG and EXPONENTIAL examples that use rules from the Table of Integrals, as well as trig identities, skip to 11:36. Nancy formerly of MathBFF explains the steps.

For my video on how to integrate using U-SUBSTITUTION, jump to: https://youtu.be/8B31SAk1nD8

For my video on how to do INTEGRATION BY PARTS, jump to: https://youtu.be/KKg88oSUv0o

1) POWER RULE: If you're integrating a polynomial, or just a power of x, you can use the Power Rule on each x-term in the polynomial. The Power Rule says that for a term that's just a power of x, such as x^3, you can integrate by raising the power by 1 AND dividing by that number (the new number you got by increasing the power by 1). For example, the integral of x^3 would be (x^4)/4. If there was a constant multiplied in front of the x-power, you can keep the constant and integrate the rest of the term (Constant Multiple Rule). For example, the integral of 6x^2 would be 6 times (x^3)/3, which simplifies to 2x^3. You can repeat these steps to integrate every term in the polynomial and string them together for the full integral. You can also keep the addition and subtraction between the terms (because of the Sum and Difference Rules). NOTE: The anti-derivative of a constant (just a number) is always that number times x (the Constant Rule), so the integral of 1 is 1*x, or x. At the end, it's very important to remember to add a constant of integration, to include a "+ c" at the end of your answer, when it is an indefinite integral (integral with no limits). ANOTHER NOTE: The Power Rule only works when the power is not -1.

2) NEGATIVE & FRACTIONAL POWERS, and RADICALS: For negative powers, you can still use the Power Rule, as long as the power is not -1. For fractional powers, you can also use the Power Rule. When you increase a fractional power by 1, you will have to simplify the power (before integrating) by getting a common denominator. For roots, like the square root of x, it's best to rewrite the radical as a fractional power, and then use the Power Rule.

3) IF THE X POWER IS -1 in the integrand, either written as x^(-1) or 1/x, you have to use a special rule, the LOG RULE, that you can find on a Table of Integrals. It says that the anti-derivative of x^(-1), or (1/x) is the natural log of the absolute value of x, plus c.

4) MORE ALGEBRA: Sometimes you may need to do a bit more algebra before integrating, so that the integrand is in a form that fits a basic integration rule, like the Power Rule. If you're integrating a rational expression, you can sometimes separate the numerator, or break the fraction into separate fractions, simplify each term, and then integrate with the Power Rule. If you have a product of x expressions, you can multiply it out, or distribute the factors so that you have just a polynomial (and can use the Power Rule).

5) TRIG & EXPONENTIAL: You can find a lot of trigonometric (and exponential) integral rules in the Table of Integrals. If you don't see it in the table, you may need to use a trig identity first, to rewrite the integrand into a form that you can integrate.

For my video on how to integrate using U-SUBSTITUTION, jump to: https://youtu.be/8B31SAk1nD8

For my video on how to do INTEGRATION BY PARTS, jump to: https://youtu.be/KKg88oSUv0o

1) POWER RULE: If you're integrating a polynomial, or just a power of x, you can use the Power Rule on each x-term in the polynomial. The Power Rule says that for a term that's just a power of x, such as x^3, you can integrate by raising the power by 1 AND dividing by that number (the new number you got by increasing the power by 1). For example, the integral of x^3 would be (x^4)/4. If there was a constant multiplied in front of the x-power, you can keep the constant and integrate the rest of the term (Constant Multiple Rule). For example, the integral of 6x^2 would be 6 times (x^3)/3, which simplifies to 2x^3. You can repeat these steps to integrate every term in the polynomial and string them together for the full integral. You can also keep the addition and subtraction between the terms (because of the Sum and Difference Rules). NOTE: The anti-derivative of a constant (just a number) is always that number times x (the Constant Rule), so the integral of 1 is 1*x, or x. At the end, it's very important to remember to add a constant of integration, to include a "+ c" at the end of your answer, when it is an indefinite integral (integral with no limits). ANOTHER NOTE: The Power Rule only works when the power is not -1.

2) NEGATIVE & FRACTIONAL POWERS, and RADICALS: For negative powers, you can still use the Power Rule, as long as the power is not -1. For fractional powers, you can also use the Power Rule. When you increase a fractional power by 1, you will have to simplify the power (before integrating) by getting a common denominator. For roots, like the square root of x, it's best to rewrite the radical as a fractional power, and then use the Power Rule.

3) IF THE X POWER IS -1 in the integrand, either written as x^(-1) or 1/x, you have to use a special rule, the LOG RULE, that you can find on a Table of Integrals. It says that the anti-derivative of x^(-1), or (1/x) is the natural log of the absolute value of x, plus c.

4) MORE ALGEBRA: Sometimes you may need to do a bit more algebra before integrating, so that the integrand is in a form that fits a basic integration rule, like the Power Rule. If you're integrating a rational expression, you can sometimes separate the numerator, or break the fraction into separate fractions, simplify each term, and then integrate with the Power Rule. If you have a product of x expressions, you can multiply it out, or distribute the factors so that you have just a polynomial (and can use the Power Rule).

5) TRIG & EXPONENTIAL: You can find a lot of trigonometric (and exponential) integral rules in the Table of Integrals. If you don't see it in the table, you may need to use a trig identity first, to rewrite the integrand into a form that you can integrate.

MIT grad shows how to integrate by parts and the LIATE trick. To skip ahead: 1) For how to use integration by parts and a good RULE OF THUMB for CHOOSING U and DV, skip to time 2:46. 2) For the TRICK FOR CHOOSING U and DV (the LIATE memory trick) skip to 10:12. Nancy formerly of MathBFF explains the steps.

WHEN to use INTEGRATION BY PARTS: If you have an integral to evaluate, and you don't already know how to integrate it, as is, then see if you can simplify it somehow with algebra. If not, try doing a substitution, like U-Substitution. If U-Substitution does not help, then you may need to use the INTEGRATION BY PARTS formula.

RULE OF THUMB: The first step to use Integration by Parts is to pick your "u" and "dv". As a general rule of thumb, whichever factor in your integrand gets simpler when you take the derivative of it, make that your u. Then make the other factor your dv (and include the dx in this dv). Note: This works if the part you chose for dv does not get any bigger or more complicated when you integrate it. After you've picked u and dv, then find du by differentiating u, and find v by integrating dv. Finally, plug everything into the integration by parts formula and simplify.

TRICK: You can instead use an acronym memory trick to choose the right "u". The acronym LIATE stands for Logs, Inverse trig functions, Algebraic functions, Trig functions, and Exponentials (in that order). If you follow those letters, LIATE, in sequence, whichever type of function you find first that you have, make that your u. Then, make the next thing you find your dv. Once you've picked u and dv this way, the steps are the same as above. Find du by differentiating u, and find v by integrating dv. Then use the IBP (integration by parts) formula and simplify for the answer.

WHEN to use INTEGRATION BY PARTS: If you have an integral to evaluate, and you don't already know how to integrate it, as is, then see if you can simplify it somehow with algebra. If not, try doing a substitution, like U-Substitution. If U-Substitution does not help, then you may need to use the INTEGRATION BY PARTS formula.

RULE OF THUMB: The first step to use Integration by Parts is to pick your "u" and "dv". As a general rule of thumb, whichever factor in your integrand gets simpler when you take the derivative of it, make that your u. Then make the other factor your dv (and include the dx in this dv). Note: This works if the part you chose for dv does not get any bigger or more complicated when you integrate it. After you've picked u and dv, then find du by differentiating u, and find v by integrating dv. Finally, plug everything into the integration by parts formula and simplify.

TRICK: You can instead use an acronym memory trick to choose the right "u". The acronym LIATE stands for Logs, Inverse trig functions, Algebraic functions, Trig functions, and Exponentials (in that order). If you follow those letters, LIATE, in sequence, whichever type of function you find first that you have, make that your u. Then, make the next thing you find your dv. Once you've picked u and dv this way, the steps are the same as above. Find du by differentiating u, and find v by integrating dv. Then use the IBP (integration by parts) formula and simplify for the answer.

MIT grad shows how to solve log equations, using LOG PROPERTIES to simplify and solve. To skip ahead: 1) For solving BASIC LOG EQUATIONS, skip to 0:22. 2) For ONE LOG ON EACH SIDE (EQUALITY property), skip to 2:36. 3) For TWO LOGS ADDED together and equaling a number (PRODUCT property), skip to 5:25. 4) For MANY LOGS, SUBTRACTED and ADDED, and using the QUOTIENT property, POWER property, and the other properties of logarithms (log rules), skip to 9:53. This video focuses on solving logarithms in equations and explains how to check the solution for an extraneous solution.

For my video on logarithm basics like how to EVALUATE LOGS, including natural logs (ln x), jump to: https://youtu.be/Zw5t6BTQYRU

SOLVING LOGARITHMIC EQUATIONS for x: Every logarithm is connected to an exponential form. The best way to figure out a log function is to REARRANGE THE LOG INTO EXPONENTIAL FORM and then solve for x. I show a quick recap of how to rewrite a log into exponential form, but for a longer explanation of rearranging into exponential form to evaluate a log, jump to my video "Logarithms... How?" (https://youtu.be/Zw5t6BTQYRU) for help with converting log to exponential form. In this video, I show you more complicated log equations with more than one log, where you'll need the LOG PROPERTIES (logarithm rules) to simplify and be able to condense down to one logarithm on a side so that you can solve.

Solving logarithmic equations with LOGS ON BOTH SIDES: If you have one log on each side of the equation, with the same base, you can use the EQUALITY PROPERTY log formula.

If you have more than one log on a side of the equation (with the same base), you can use the PRODUCT PROPERTY (if the logs are added) or the QUOTIENT PROPERTY (if the logs are subtracted) to combine the logs into one log. If you have a number multiplied in front of a log, as a coefficient, you can use the POWER PROPERTY logarithm formula to bring the coefficient up as a power inside the log argument. Once you simplify the log equation using the product, quotient, and/or power properties, you can either 1) solve by using the equality property if you just have one log on each side (with same base), or 2) solve by rewriting into exponential form if you have a log on one side and a number on the other side. IMPORTANT - CHECK SOLUTIONS: When solving logarithmic functions, you have to CHECK your solutions for log equations. Any numbers you get as solutions you have to plug back into the original equation to check. If a solution does not work or gives an undefined value (for example: log of a negative number), then it is an "extraneous solution" and must be thrown out!

For my video on logarithm basics like how to EVALUATE LOGS, including natural logs (ln x), jump to: https://youtu.be/Zw5t6BTQYRU

SOLVING LOGARITHMIC EQUATIONS for x: Every logarithm is connected to an exponential form. The best way to figure out a log function is to REARRANGE THE LOG INTO EXPONENTIAL FORM and then solve for x. I show a quick recap of how to rewrite a log into exponential form, but for a longer explanation of rearranging into exponential form to evaluate a log, jump to my video "Logarithms... How?" (https://youtu.be/Zw5t6BTQYRU) for help with converting log to exponential form. In this video, I show you more complicated log equations with more than one log, where you'll need the LOG PROPERTIES (logarithm rules) to simplify and be able to condense down to one logarithm on a side so that you can solve.

Solving logarithmic equations with LOGS ON BOTH SIDES: If you have one log on each side of the equation, with the same base, you can use the EQUALITY PROPERTY log formula.

If you have more than one log on a side of the equation (with the same base), you can use the PRODUCT PROPERTY (if the logs are added) or the QUOTIENT PROPERTY (if the logs are subtracted) to combine the logs into one log. If you have a number multiplied in front of a log, as a coefficient, you can use the POWER PROPERTY logarithm formula to bring the coefficient up as a power inside the log argument. Once you simplify the log equation using the product, quotient, and/or power properties, you can either 1) solve by using the equality property if you just have one log on each side (with same base), or 2) solve by rewriting into exponential form if you have a log on one side and a number on the other side. IMPORTANT - CHECK SOLUTIONS: When solving logarithmic functions, you have to CHECK your solutions for log equations. Any numbers you get as solutions you have to plug back into the original equation to check. If a solution does not work or gives an undefined value (for example: log of a negative number), then it is an "extraneous solution" and must be thrown out!

MIT grad shows the DEFINITION of the derivative and how to FIND the derivative using that limit definition. To skip ahead: 1) For what the derivative MEANS, skip to 0:23. 2) For the SLOPE OF THE SECANT line, skip to 2:53. 3) For the SECANT BECOMING THE TANGENT and DEFINITION OF THE DERIVATIVE, skip to 5:54. 4) For how to FIND THE DERIVATIVE USING THE (LIMIT) DEFINITION of the derivative, skip to 10:28. Nancy formerly of MathBFF explains the steps.

For my video on the shorter, faster DERIVATIVE RULES for how to take a derivative, jump to: https://youtu.be/QqF3i1pnyzU

INTRODUCTION to derivatives:

1) WHAT IS THE DERIVATIVE? It's a function that tells you the slope (of the line tangent to the curve) at every point. Another way to think of it is that the derivative gives you the rate of change at any instant (the "instantaneous rate of change" at each point). You can find the derivative either with the proper DEFINITION OF THE DERIVATIVE ("by the limit process") or the faster, simpler way with the shortcut derivative rules such as the Power Rule, Product Rule, Quotient Rule, and Chain Rule. This video shows the first way, with the DEFINITION of the derivative, and how to use it to calculate the derivative.

2) SLOPE OF SECANT: To make the definition of the derivative, we can start with the slope of the straight secant line through a point x and some other point nearby, h distance away horizontally. If we label the two points and use the slope formula to write an expression for the slope of this secant line, the expression you get is [f(x+h) - f(x)] / h, which is also known as the DIFFERENCE QUOTIENT.

3) DEFINITION OF DERIVATIVE: That straight line slope we just found is actually a decent ESTIMATE for the slope at x, but it's not really the slope at x. We can get the EXACT slope of the tangent line at x by closing in on x and narrowing h to zero, by taking the LIMIT of the secant slope, as h approaches 0. The slope of the secant line becomes the slope of the tangent line. Not only is this limit equal to the slope of the tangent line, it is the definition of the derivative (when the limit exists) or f'(x).

4) HOW TO FIND THE DERIVATIVE OF A FUNCTION USING (LIMIT) DEFINITION: If you have to find the derivative "using the definition of the derivative" or "by the limit process", then you can use the limit definition we just found for any f(x) equation you're given. Remember that the f(x+h) part of the formula means to replace x with (x+h) anywhere x appears in the function f(x). If you want help knowing how to find the limit at the end, you can jump to my video "How to Find Any Limit": https://youtu.be/nJZm-zp639s

For an intro to the concept of limits, jump to: https://youtu.be/poBobcFn1Co

For my video on the shorter, faster DERIVATIVE RULES for how to take a derivative, jump to: https://youtu.be/QqF3i1pnyzU

INTRODUCTION to derivatives:

1) WHAT IS THE DERIVATIVE? It's a function that tells you the slope (of the line tangent to the curve) at every point. Another way to think of it is that the derivative gives you the rate of change at any instant (the "instantaneous rate of change" at each point). You can find the derivative either with the proper DEFINITION OF THE DERIVATIVE ("by the limit process") or the faster, simpler way with the shortcut derivative rules such as the Power Rule, Product Rule, Quotient Rule, and Chain Rule. This video shows the first way, with the DEFINITION of the derivative, and how to use it to calculate the derivative.

2) SLOPE OF SECANT: To make the definition of the derivative, we can start with the slope of the straight secant line through a point x and some other point nearby, h distance away horizontally. If we label the two points and use the slope formula to write an expression for the slope of this secant line, the expression you get is [f(x+h) - f(x)] / h, which is also known as the DIFFERENCE QUOTIENT.

3) DEFINITION OF DERIVATIVE: That straight line slope we just found is actually a decent ESTIMATE for the slope at x, but it's not really the slope at x. We can get the EXACT slope of the tangent line at x by closing in on x and narrowing h to zero, by taking the LIMIT of the secant slope, as h approaches 0. The slope of the secant line becomes the slope of the tangent line. Not only is this limit equal to the slope of the tangent line, it is the definition of the derivative (when the limit exists) or f'(x).

4) HOW TO FIND THE DERIVATIVE OF A FUNCTION USING (LIMIT) DEFINITION: If you have to find the derivative "using the definition of the derivative" or "by the limit process", then you can use the limit definition we just found for any f(x) equation you're given. Remember that the f(x+h) part of the formula means to replace x with (x+h) anywhere x appears in the function f(x). If you want help knowing how to find the limit at the end, you can jump to my video "How to Find Any Limit": https://youtu.be/nJZm-zp639s

For an intro to the concept of limits, jump to: https://youtu.be/poBobcFn1Co

MIT grad shows how to use the chain rule to find the derivative and WHEN to use it. To skip ahead: 1) For how to use the CHAIN RULE or "OUTSIDE-INSIDE rule", skip to time 0:17. 1b) For how to know WHEN YOU NEED the chain rule, skip to 4:35. 2) For another example with the POWER RULE in the chain rule, skip to 7:05. 3) For a TRIG derivative chain rule example, skip to 9:33. 3b) For the formal chain rule FORMULA, skip to 11:36. PS) For a DOUBLE CHAIN RULE (or "repeated use of the chain rule") example, skip to 13:33. Nancy formerly of MathBFF explains the steps.

1) The CHAIN RULE is one of the derivative rules. You need it to take the derivative when you have a function inside a function, or a "composite function". For ex, in the equation y = (3x + 1)^7, since the function 3x+1 is inside a larger, outer function, the power of 7, you'll need the chain rule to find the correct derivative. How do you use the chain rule? You can think of it as the "OUTSIDE-INSIDE" rule: take the DERIVATIVE of JUST the OUTSIDE function first, LEAVING THE INSIDE FUNCTION alone (unchanged), then MULTIPLY BY the DERIVATIVE of JUST the INSIDE function. Sometimes you might hear this expressed as: take the derivative of the outer function, "evaluated at the inner function", times the derivative of just the inner function. For our ex, first take the derivative of the outer function (the power of 7) to get 7*(3x + 1)^6 since the derivative "power rule" tells you to bring down the power to the front (as a constant or coefficient just multiplied in the front) and then decrease the power by 1, which leaves a power of 6. Notice that you leave the inside function the way it is and just rewrite it for now. Then you multiply by the derivative of just the inner function, 3x + 1. Since the derivative of 3x + 1 is just 3, the full derivative (dy/dx) is: 7*[(3x + 1)^6]*3, which is just 21(3x + 1)^6.

1b) HOW do you know WHEN TO USE the chain rule? If the original equation had just been x^7, there would be no need for the chain rule. It's when you have something more than just x inside that you should use the chain rule, such as (3x + 1)^7 or even (x^2 + 1)^7. Sometimes the chain rule may make no difference. For instance, if you have the function (x + 1)^7, taking the derivative of the inside function just gives you 1, so multiplying by that inside derivative of 1 will not change the overall answer. However, it can't hurt to use the chain rule anyway, so it's a good idea to get in the habit of using it so that you don't forget it when it really does make a difference.

2) Another chain POWER RULE example: To find the derivative of h(x) = (x^2 + 5x - 6)^9, use the same steps as above to first take the outside derivative and then multiply by the inside derivative. In this case, the derivative, dh/dx (or h'(x)) is equal to 9(x^2 + 5x - 6)^8 * (2x + 5). Using the chain rule with the power rule is sometimes called the "power chain rule".

3) TRIG EXAMPLE: the idea is the same as above even if you're using the chain rule to differentiate something like a trigonometric function. If you have anything more than just x inside the trig function, you'll need the chain rule to find the derivative. For the equation y = sin(x^2 - 3x), you first take the derivative of the outer function, just the sine function. Since the derivative of sine is cosine, the outside derivative (with the inside left unchanged) is cos(x^2 - 3x). Then, find the derivative of just the inside (of just the x^2 - 3x part), and multiply by that. Since the derivative of x^2 - 3x is 2x - 3, the full derivative answer is dy/dx = cos(x^2 - 3x)*(2x - 3).

3b) FORMULA: Although it's easier to think about the chain rule as the "outside-inside rule", if for any reason you have to use the formal chain rule formula, check out the two versions I show here. Both are based on the equation being a composition of functions, f(g(x)). The second version shown uses Liebniz notation. Either way, both show a component of the derivative that comes from the inside function, and it's important not to forget to multiply by this inside derivative factor if you want to get the right full derivative answer.

P.S.) DOUBLE CHAIN RULE: Sometimes you might have to use the chain rule more than once, known as "repeated use of the chain rule". In y = (1 + cos2x)^2, not only would you need to take the derivative of the outside power of 2, as well as multiply by the derivative of the inside function, 1 + cos2x, but after that you would ALSO then need to multiply by the derivative of the 2x inside cosine because that inside function was 1 + cos2x and not just 1 + cosx. This means you would use the chain rule twice. The idea is that you have to keep taking the derivative of the inner functions until you have reached every inner function that is more complicated than just "x".

1) The CHAIN RULE is one of the derivative rules. You need it to take the derivative when you have a function inside a function, or a "composite function". For ex, in the equation y = (3x + 1)^7, since the function 3x+1 is inside a larger, outer function, the power of 7, you'll need the chain rule to find the correct derivative. How do you use the chain rule? You can think of it as the "OUTSIDE-INSIDE" rule: take the DERIVATIVE of JUST the OUTSIDE function first, LEAVING THE INSIDE FUNCTION alone (unchanged), then MULTIPLY BY the DERIVATIVE of JUST the INSIDE function. Sometimes you might hear this expressed as: take the derivative of the outer function, "evaluated at the inner function", times the derivative of just the inner function. For our ex, first take the derivative of the outer function (the power of 7) to get 7*(3x + 1)^6 since the derivative "power rule" tells you to bring down the power to the front (as a constant or coefficient just multiplied in the front) and then decrease the power by 1, which leaves a power of 6. Notice that you leave the inside function the way it is and just rewrite it for now. Then you multiply by the derivative of just the inner function, 3x + 1. Since the derivative of 3x + 1 is just 3, the full derivative (dy/dx) is: 7*[(3x + 1)^6]*3, which is just 21(3x + 1)^6.

1b) HOW do you know WHEN TO USE the chain rule? If the original equation had just been x^7, there would be no need for the chain rule. It's when you have something more than just x inside that you should use the chain rule, such as (3x + 1)^7 or even (x^2 + 1)^7. Sometimes the chain rule may make no difference. For instance, if you have the function (x + 1)^7, taking the derivative of the inside function just gives you 1, so multiplying by that inside derivative of 1 will not change the overall answer. However, it can't hurt to use the chain rule anyway, so it's a good idea to get in the habit of using it so that you don't forget it when it really does make a difference.

2) Another chain POWER RULE example: To find the derivative of h(x) = (x^2 + 5x - 6)^9, use the same steps as above to first take the outside derivative and then multiply by the inside derivative. In this case, the derivative, dh/dx (or h'(x)) is equal to 9(x^2 + 5x - 6)^8 * (2x + 5). Using the chain rule with the power rule is sometimes called the "power chain rule".

3) TRIG EXAMPLE: the idea is the same as above even if you're using the chain rule to differentiate something like a trigonometric function. If you have anything more than just x inside the trig function, you'll need the chain rule to find the derivative. For the equation y = sin(x^2 - 3x), you first take the derivative of the outer function, just the sine function. Since the derivative of sine is cosine, the outside derivative (with the inside left unchanged) is cos(x^2 - 3x). Then, find the derivative of just the inside (of just the x^2 - 3x part), and multiply by that. Since the derivative of x^2 - 3x is 2x - 3, the full derivative answer is dy/dx = cos(x^2 - 3x)*(2x - 3).

3b) FORMULA: Although it's easier to think about the chain rule as the "outside-inside rule", if for any reason you have to use the formal chain rule formula, check out the two versions I show here. Both are based on the equation being a composition of functions, f(g(x)). The second version shown uses Liebniz notation. Either way, both show a component of the derivative that comes from the inside function, and it's important not to forget to multiply by this inside derivative factor if you want to get the right full derivative answer.

P.S.) DOUBLE CHAIN RULE: Sometimes you might have to use the chain rule more than once, known as "repeated use of the chain rule". In y = (1 + cos2x)^2, not only would you need to take the derivative of the outside power of 2, as well as multiply by the derivative of the inside function, 1 + cos2x, but after that you would ALSO then need to multiply by the derivative of the 2x inside cosine because that inside function was 1 + cos2x and not just 1 + cosx. This means you would use the chain rule twice. The idea is that you have to keep taking the derivative of the inner functions until you have reached every inner function that is more complicated than just "x".

MIT grad introduces logs and shows how to evaluate them. To skip ahead: 1) For how to understand and evaluate BASIC LOGS, skip to time 0:52. 2) For how to evaluate weirder logs, including the log of 0, 1, a FRACTION, or a NEGATIVE number, skip to time 6:44. 3) For NATURAL LOGS (LN X), skip to time 11:17. 4) For even weirder logs, including SOLVING for X and using the CHANGE-OF-BASE formula, skip to time 14:56. Nancy formerly of MathBFF explains the steps.

1) BASIC LOGS: you can read log notation as "log, base 3, of 9 equals X". The small (subscript) number is called the base. You can always evaluate a log expression by rearranging it into something called exponential form. Every log expression is connected to an exponential expression. In this example, the log is connected to the exponential form "3 to the X power equals 9". This means, "3 raised to what power gives you 9?" Since 3 raised to the power of 2 equals 9, the answer for X is 2. This is also the answer for the value of the log expression. The log is always equal to the power (or exponent) in the exponential version, and in this case it equals 2. If you want, you can find the log value in your head just by asking yourself what power you need in order to turn the base number into the middle number ("argument" number). Note: if there is no base number in the log expression (no little subscript number), then the base is 10, since 10 is the default base.

2) WEIRDER LOGS (log of 0, 1, a negative number, or a fraction): you can use the same steps to rearrange log expressions that have a fraction, negative number, 0, or 1 in them. You can still rearrange them to be in exponential form just like you can with the basic logs from earlier. The log of 1 will always be 0, since 0 is the only power that can turn a base into 1. The log of 0 will always be undefined, since no power can turn a base into 0. The log of a negative number is undefined in the real number system, since no real power can turn a positive base into a negative number.

3) NATURAL LOGS (ln x): the natural log is just a special type of log where the base is e (the special math constant e, which is approximately 2.718 if you plug it into your calculator). You can use the same steps for rearranging the log expression into exponential form. Just remember that ln x means log, base e.

4) EVEN WEIRDER LOGS (solving for X, change-of-base formula): even if there is an X variable in the log part of an equation, you can still rearrange the equation into exponential form. This will let you solve for X. Sometimes you might need to use the change-of-base formula to evaluate a log expression. If there is no whole number power you know that works, it may actually be a decimal power that you can find by using the change-of-base formula. For example, you can re-write log, base 2, of 7 as (log 7)/(log 2) and use your calculator to find the decimal number if you need it.

1) BASIC LOGS: you can read log notation as "log, base 3, of 9 equals X". The small (subscript) number is called the base. You can always evaluate a log expression by rearranging it into something called exponential form. Every log expression is connected to an exponential expression. In this example, the log is connected to the exponential form "3 to the X power equals 9". This means, "3 raised to what power gives you 9?" Since 3 raised to the power of 2 equals 9, the answer for X is 2. This is also the answer for the value of the log expression. The log is always equal to the power (or exponent) in the exponential version, and in this case it equals 2. If you want, you can find the log value in your head just by asking yourself what power you need in order to turn the base number into the middle number ("argument" number). Note: if there is no base number in the log expression (no little subscript number), then the base is 10, since 10 is the default base.

2) WEIRDER LOGS (log of 0, 1, a negative number, or a fraction): you can use the same steps to rearrange log expressions that have a fraction, negative number, 0, or 1 in them. You can still rearrange them to be in exponential form just like you can with the basic logs from earlier. The log of 1 will always be 0, since 0 is the only power that can turn a base into 1. The log of 0 will always be undefined, since no power can turn a base into 0. The log of a negative number is undefined in the real number system, since no real power can turn a positive base into a negative number.

3) NATURAL LOGS (ln x): the natural log is just a special type of log where the base is e (the special math constant e, which is approximately 2.718 if you plug it into your calculator). You can use the same steps for rearranging the log expression into exponential form. Just remember that ln x means log, base e.

4) EVEN WEIRDER LOGS (solving for X, change-of-base formula): even if there is an X variable in the log part of an equation, you can still rearrange the equation into exponential form. This will let you solve for X. Sometimes you might need to use the change-of-base formula to evaluate a log expression. If there is no whole number power you know that works, it may actually be a decimal power that you can find by using the change-of-base formula. For example, you can re-write log, base 2, of 7 as (log 7)/(log 2) and use your calculator to find the decimal number if you need it.

MIT grad shows what a limit is, how to read the notation, what it means on a graph and how to find the limit on a graph. To skip ahead: 1) For how to understand limit NOTATION and the CONCEPT of the limit, skip to time 0:34. 2) For WHICH WAY TO LOOK AT THE GRAPH to find the limit, including when to use the X and when to use the Y, skip to time 1:52. 3) For ONE-SIDED LIMITS notation, including the LEFT-SIDED LIMIT and RIGHT-SIDED LIMIT, skip to time 7:54. 4) For how to understand limits where X APPROACHES INFINITY or negative infinity, skip to time 10:24. Nancy formerly of MathBFF explains the steps.

For HOW TO FIND THE LIMIT (at a finite value), jump to https://youtu.be/nJZm-zp639s

For HOW TO FIND THE LIMIT AT INFINITY, jump to https://youtu.be/nViVR1rImUE

1) LIMIT NOTATION and WHAT A LIMIT MEANS: You can read the limit notation as "the limit, as x approaches 1, of f(x)". This means "when x gets very close to 1, what number is y getting very close to?" The limit is always equal to a y-value. It is a way of predicting what y-value we would expect to have, if we tend toward a specific x-value. Why do we need the limit? One reason is that there are sometimes "blindspots" such as gaps (holes) in a function in which we cannot see what the function is doing exactly at a point, but we can see what it is doing as we head toward that point.

2) HOW TO LOOK AT THE GRAPH to find the limit: a) For a removable discontinuity (hole), b) For a removable discontinuity with a point defined above, and c) For a normal line. When you're finding an overall limit, the hidden, implied meaning is that YOU MUST CHECK BOTH SIDES OF THE X-VALUE, from the left and from the right. If both sides give you the same limit value, then that value is your overall limit. In our example, to find the limit from the left side, TRACE X VALUES from the left of 1 but headed toward 1 (the actual motion is to the right), and check to SEE WHAT Y-VALUE the function is tending toward. That y-value is the left-hand limit. To find the limit from the right side, trace x values from the right of 1 but headed toward 1 (the actual motion is to the left), and again check to see what Y-VALUE the function is heading toward. That y-value is the right-hand limit. Since the left limit (2) and the right limit (2) are the same in our example, the overall limit answer is 2. If they were not the same, we could not give a limit value (see #3). IMPORTANT TAKEAWAY: For the limit, we DO NOT CARE what is happening EXACTLY AT THE X-VALUE and ONLY CARE what y-values the function is hitting NEAR the x-value, as we get closer and closer to that x. In other words, the limit, as x approaches 1, of f(x) can equal 2, even if (1) = 3 or some other number different from 2, or even if f(1) is not defined or indeterminate.

3) ONE-SIDED LIMITS (RIGHT-SIDED LIMIT and LEFT-SIDED LIMIT) for a jump discontinuity: as you saw in #2, to find the overall limit, you have to check both the left and right limits. Sometimes the left limit and right limit are not the same. If you get a limit question with notation in which the x is approaching a number but with a plus sign or minus sign as a superscript, that is notation for a one-sided limit. The minus sign means the limit from the left, and the plus sign means the limit from the right. IF THE LEFT limit AND RIGHT limit are NOT THE SAME, then the overall limit DOES NOT EXIST (sometimes written as "DNE"). Even if the left and right limits are different, you can still write the left-sided limit and right-sided limit values separately.

4) LIMITS in which X APPROACHES INFINITY (or negative infinity): Another "blindspot" is when x goes toward infinity or negative infinity. Since we can never "see" exactly at infinity (or negative infinity), we can use the idea of the limit to say what y-value it looks like the function is headed toward when our x value approaches infinity. If x is approaching INFINITY, TRACE x values TOWARD THE RIGHT (the large positive direction) on the graph, and see what y-value the function is approaching. That y-value is the limit. Note that the function may be approaching an asymptote. If x is approaching NEGATIVE INFINITY, trace x values TOWARD THE LEFT (the large negative direction), and check what y-value the function is getting closer and closer to on the graph. That y-value is the limit.

For HOW TO FIND THE LIMIT (at a finite value), jump to https://youtu.be/nJZm-zp639s

For HOW TO FIND THE LIMIT AT INFINITY, jump to https://youtu.be/nViVR1rImUE

1) LIMIT NOTATION and WHAT A LIMIT MEANS: You can read the limit notation as "the limit, as x approaches 1, of f(x)". This means "when x gets very close to 1, what number is y getting very close to?" The limit is always equal to a y-value. It is a way of predicting what y-value we would expect to have, if we tend toward a specific x-value. Why do we need the limit? One reason is that there are sometimes "blindspots" such as gaps (holes) in a function in which we cannot see what the function is doing exactly at a point, but we can see what it is doing as we head toward that point.

2) HOW TO LOOK AT THE GRAPH to find the limit: a) For a removable discontinuity (hole), b) For a removable discontinuity with a point defined above, and c) For a normal line. When you're finding an overall limit, the hidden, implied meaning is that YOU MUST CHECK BOTH SIDES OF THE X-VALUE, from the left and from the right. If both sides give you the same limit value, then that value is your overall limit. In our example, to find the limit from the left side, TRACE X VALUES from the left of 1 but headed toward 1 (the actual motion is to the right), and check to SEE WHAT Y-VALUE the function is tending toward. That y-value is the left-hand limit. To find the limit from the right side, trace x values from the right of 1 but headed toward 1 (the actual motion is to the left), and again check to see what Y-VALUE the function is heading toward. That y-value is the right-hand limit. Since the left limit (2) and the right limit (2) are the same in our example, the overall limit answer is 2. If they were not the same, we could not give a limit value (see #3). IMPORTANT TAKEAWAY: For the limit, we DO NOT CARE what is happening EXACTLY AT THE X-VALUE and ONLY CARE what y-values the function is hitting NEAR the x-value, as we get closer and closer to that x. In other words, the limit, as x approaches 1, of f(x) can equal 2, even if (1) = 3 or some other number different from 2, or even if f(1) is not defined or indeterminate.

3) ONE-SIDED LIMITS (RIGHT-SIDED LIMIT and LEFT-SIDED LIMIT) for a jump discontinuity: as you saw in #2, to find the overall limit, you have to check both the left and right limits. Sometimes the left limit and right limit are not the same. If you get a limit question with notation in which the x is approaching a number but with a plus sign or minus sign as a superscript, that is notation for a one-sided limit. The minus sign means the limit from the left, and the plus sign means the limit from the right. IF THE LEFT limit AND RIGHT limit are NOT THE SAME, then the overall limit DOES NOT EXIST (sometimes written as "DNE"). Even if the left and right limits are different, you can still write the left-sided limit and right-sided limit values separately.

4) LIMITS in which X APPROACHES INFINITY (or negative infinity): Another "blindspot" is when x goes toward infinity or negative infinity. Since we can never "see" exactly at infinity (or negative infinity), we can use the idea of the limit to say what y-value it looks like the function is headed toward when our x value approaches infinity. If x is approaching INFINITY, TRACE x values TOWARD THE RIGHT (the large positive direction) on the graph, and see what y-value the function is approaching. That y-value is the limit. Note that the function may be approaching an asymptote. If x is approaching NEGATIVE INFINITY, trace x values TOWARD THE LEFT (the large negative direction), and check what y-value the function is getting closer and closer to on the graph. That y-value is the limit.

MIT grad shows how to do integration using u-substitution (Calculus). To skip ahead: 1) for a BASIC example where your du gives you exactly the expression you need in order to substitute, skip to time 1:30. 2) For an example where you have to REARRANGE THE DU by multiplying or dividing because the du has a different number or sign than what you need, skip to time 8:21. 3) For one where you have to REARRANGE THE U by subtracting or adding because the du expression cannot give you the expression you need, skip to 17:15. 4) For u-substitution with TRIG (SIN/COS) and the power rule, skip to 22:35. Nancy formerly of MathBFF explains the steps.

For my video on the BASICS of INTEGRATION, jump to: https://youtu.be/e1nxhJQyLYI

With all u-substitution integration problems:

The FIRST STEP is to pick your "u". The best choice is usually the longer x-expression that is inside a power or a square root or the denominator, etc (in an "inside function"). Set u equal to this x-expression.

The SECOND STEP is to find "du" by taking the derivative of the u expression with respect to x. For instance, if you have u=3x+2, your du would then be du=3dx. **NOTE: Remember to include "dx" at the end of your du differential expression.

The THIRD STEP is to substitute u and du into the integral everywhere in place of x and dx. **NOTE: if your du does not perfectly match what you need in order to completely substitute before integrating, you must rearrange the du, or sometimes rearrange the u, in order to fully substitute before integrating. For an example of each, see example #2 (time: 8:21) and example #3 (time: 17:15).

The FOURTH STEP is to integrate, remembering to add "+ C" at the end since you integrated an indefinite integral (no limits). The du goes away when you integrate.

The LAST STEP is to "back-substitute" by replacing everywhere u appears with the x-expression that you chose u to be.

For my video on the BASICS of INTEGRATION, jump to: https://youtu.be/e1nxhJQyLYI

With all u-substitution integration problems:

The FIRST STEP is to pick your "u". The best choice is usually the longer x-expression that is inside a power or a square root or the denominator, etc (in an "inside function"). Set u equal to this x-expression.

The SECOND STEP is to find "du" by taking the derivative of the u expression with respect to x. For instance, if you have u=3x+2, your du would then be du=3dx. **NOTE: Remember to include "dx" at the end of your du differential expression.

The THIRD STEP is to substitute u and du into the integral everywhere in place of x and dx. **NOTE: if your du does not perfectly match what you need in order to completely substitute before integrating, you must rearrange the du, or sometimes rearrange the u, in order to fully substitute before integrating. For an example of each, see example #2 (time: 8:21) and example #3 (time: 17:15).

The FOURTH STEP is to integrate, remembering to add "+ C" at the end since you integrated an indefinite integral (no limits). The du goes away when you integrate.

The LAST STEP is to "back-substitute" by replacing everywhere u appears with the x-expression that you chose u to be.

MIT grad shows the easiest way to complete the square to solve a quadratic equation. To skip ahead: 1) for a quadratic that STARTS WITH X^2, skip to time 1:42. 2) For a quadratic that STARTS WITH 2X^2, 3X^2, etc., skip to time 6:46. 3) For NEGATIVE leading term like -X^2, skip to 13:34. 4) If there's NO X TERM (ex. 3x^2 - 121 = 0), then you cannot complete the square and can solve directly - skip to 16:30. Nancy formerly of MathBFF explains the steps.

How to complete the square when solving quadratic equations:

Completing the square means re-arranging the quadratic equation into the neat, "perfect square" form (x + number)^2 equals another number, or (x - number)^2 equals another number. The point of doing this is so that you can solve by just square-rooting both sides of the equation.

If your quadratic equation starts with...

1) just X^2 (for ex: x^2 + 6x - 7 = 0), move constant to right, add (6/2)^2 to both sides, write left side as perfect square, and square root both sides to solve. Remember you get PLUS and MINUS solutions when you square root the constant side.

2) 2X^2 or 3X^2 or 4X^2 etc. (for ex: 2x^2 - 10x - 3 = 0), move constant to right side, DIVIDE EVERY TERM by leading coefficient 2 on left and right side, add (5/2)^2 to both sides, write left side as perfect square and square root both sides to solve.

3) -X^2 or any negative coefficient (for ex: -x^2 - 6x + 7 = 0), move constant to right side, DIVIDE EVERY TERM by -1 on left and right side. (NOTE: if you had -2x^2 in your equation, you would divide every term by -2 on left and right side). Then add (6/2)^2 to both sides, write left side as perfect square and square root both sides to solve.

If you don't have an X term in equation, (for ex: 3x^2 - 121 = 0), then you cannot complete the square. Just move the constant to the right, divide both sides by 3, and square root both sides to solve. Remember plus and minus solutions on the right.

How to complete the square when solving quadratic equations:

Completing the square means re-arranging the quadratic equation into the neat, "perfect square" form (x + number)^2 equals another number, or (x - number)^2 equals another number. The point of doing this is so that you can solve by just square-rooting both sides of the equation.

If your quadratic equation starts with...

1) just X^2 (for ex: x^2 + 6x - 7 = 0), move constant to right, add (6/2)^2 to both sides, write left side as perfect square, and square root both sides to solve. Remember you get PLUS and MINUS solutions when you square root the constant side.

2) 2X^2 or 3X^2 or 4X^2 etc. (for ex: 2x^2 - 10x - 3 = 0), move constant to right side, DIVIDE EVERY TERM by leading coefficient 2 on left and right side, add (5/2)^2 to both sides, write left side as perfect square and square root both sides to solve.

3) -X^2 or any negative coefficient (for ex: -x^2 - 6x + 7 = 0), move constant to right side, DIVIDE EVERY TERM by -1 on left and right side. (NOTE: if you had -2x^2 in your equation, you would divide every term by -2 on left and right side). Then add (6/2)^2 to both sides, write left side as perfect square and square root both sides to solve.

If you don't have an X term in equation, (for ex: 3x^2 - 121 = 0), then you cannot complete the square. Just move the constant to the right, divide both sides by 3, and square root both sides to solve. Remember plus and minus solutions on the right.

MIT grad shows how to do implicit differentiation to find dy/dx (Calculus). To skip ahead: 1) For a BASIC example using the POWER RULE, skip to time 3:57. 2) For an example that uses a TRIG FUNCTION and the PRODUCT RULE, skip to time 7:20. Nancy formerly of MathBFF explains the steps.

What is implicit differentiation? Up until now, most functions you've had to differentiate have probably been written EXPLICITLY as a function of x, such as y = x^2. However, if a function y is written IMPLICITLY as a function of x, such as x^2 + y^2 = 9, you will need to use implicit differentiation to find the derivative dy/dx. The only difference is that any time you take the derivative of y, you must also multiply by dy/dx. The reason is that y is dependent on x. You can think of y as containing some x expression inside it, so

implicit differentiation is basically a special instance of the chain rule in which you must take the outside derivative but also multiply by the inside derivative, which is dy/dx in this case.

Here are the steps to doing implicit differentiation to find DY/DX:

1) TAKE THE DERIVATIVE OF BOTH SIDES, MULTIPLYING BY DY/DX every time you take the derivative of a Y: The first step is to take the derivative of both sides of the equation, with respect to x, but to attach a dy/dx if you ever take the derivative of y. For instance, to implicitly differentiate the equation x^2 + y^2 = 9, take the derivative of both sides with respect to x. On the left, the derivative of the x^2 term is just 2x. To differentiate the y^2 term, in this case first use the power rule to get 2y, and THEN, because y is dependent on x, you must multiply the term by dy/dx so that you have y^2 times dy/dx. Don't forget to differentiate the right side of the original equation as well, which was the constant 9, so the derivative is just 0. Your new, differentiated equation is then 2x plus y^2 dy/dx = 0.

2) GET DY/DX ALONE ON ONE SIDE: The third step is to solve for dy/dx, or in other words, to get dy/dx alone on one side of the equation. Next, DISTRIBUTE AND EXPAND BOTH SIDES if necessary distributing or opening up any terms that have parentheses. Then, if dy/dx appears in more than one term, get those terms together on one side and use the SIMPLIFICATION TRICK OF FACTORING out dy/dx from those terms so that it then appears only once on that side. Finally, DIVIDE OUT any factor that is currently multiplied by the dy/dx so that you are left with dy/dx alone on one side and an equation that looks like dy/dx = some other expression.

NOTE: Only multiply by dy/dx if you are taking the derivative of y. In more challenging examples like the second example in this video, you do not need to attach a dy/dx if you are just multiplying by y. For instance, when differentiating a term like xy, using the product rule gives you x times dy/dx plus y. Notice that in the second term from the product rule, it was not necessary to attach a dy/dx to the y, since you were not taking the derivative of y in that term.

What is implicit differentiation? Up until now, most functions you've had to differentiate have probably been written EXPLICITLY as a function of x, such as y = x^2. However, if a function y is written IMPLICITLY as a function of x, such as x^2 + y^2 = 9, you will need to use implicit differentiation to find the derivative dy/dx. The only difference is that any time you take the derivative of y, you must also multiply by dy/dx. The reason is that y is dependent on x. You can think of y as containing some x expression inside it, so

implicit differentiation is basically a special instance of the chain rule in which you must take the outside derivative but also multiply by the inside derivative, which is dy/dx in this case.

Here are the steps to doing implicit differentiation to find DY/DX:

1) TAKE THE DERIVATIVE OF BOTH SIDES, MULTIPLYING BY DY/DX every time you take the derivative of a Y: The first step is to take the derivative of both sides of the equation, with respect to x, but to attach a dy/dx if you ever take the derivative of y. For instance, to implicitly differentiate the equation x^2 + y^2 = 9, take the derivative of both sides with respect to x. On the left, the derivative of the x^2 term is just 2x. To differentiate the y^2 term, in this case first use the power rule to get 2y, and THEN, because y is dependent on x, you must multiply the term by dy/dx so that you have y^2 times dy/dx. Don't forget to differentiate the right side of the original equation as well, which was the constant 9, so the derivative is just 0. Your new, differentiated equation is then 2x plus y^2 dy/dx = 0.

2) GET DY/DX ALONE ON ONE SIDE: The third step is to solve for dy/dx, or in other words, to get dy/dx alone on one side of the equation. Next, DISTRIBUTE AND EXPAND BOTH SIDES if necessary distributing or opening up any terms that have parentheses. Then, if dy/dx appears in more than one term, get those terms together on one side and use the SIMPLIFICATION TRICK OF FACTORING out dy/dx from those terms so that it then appears only once on that side. Finally, DIVIDE OUT any factor that is currently multiplied by the dy/dx so that you are left with dy/dx alone on one side and an equation that looks like dy/dx = some other expression.

NOTE: Only multiply by dy/dx if you are taking the derivative of y. In more challenging examples like the second example in this video, you do not need to attach a dy/dx if you are just multiplying by y. For instance, when differentiating a term like xy, using the product rule gives you x times dy/dx plus y. Notice that in the second term from the product rule, it was not necessary to attach a dy/dx to the y, since you were not taking the derivative of y in that term.

MIT grad shows how to remember the unit circle angles and points. The cos value is the first number in the point, and the sin is the second coordinate in the point. There are patterns within the unit circle that make it easier to understand and to memorize. To skip ahead: 1) For the ANGLES on the unit circle, skip to time 1:00. 2) For how to find COS (cosine function) values, skip to 4:18. 3) For the SIN (sine function) values, skip to 7:39. 4) For the COMPLETE UNIT CIRCLE chart explained, with all the points, and also when to use a NEGATIVE SIGN in some quadrants, skip to 8:17. 5) For DEGREES of the angles in the trig unit circle, skip to 10:50. This video gives memorization help for the unit circle with a trick, as well as a review of how to use the circle to find sin, cos, and tan. The cosine and sine values are the x and y coordinates of the unit circle points.

Unit circle trigonometry comes up a lot in geometry, precalculus, and even calculus problems. The trig unit circle is a circle of radius one. For each angle, there is a point on the trig circle whose x-coordinate is the cosine value of the angle and whose y-coordinate is the sine value of the angle. It is a way of finding exact values of trig functions. Here is how to memorize the unit circle values, how to fill in the unit circle, and how to use the unit circle:

ANGLES (in radians):

First, the radian angle measures of the four "corner" points on the radian circle are 0, pi/2, pi, 3pi/2, and 2pi. The 2pi angle is one complete full circle around the unit circle (radians) and is in the same position as the 0 angle measure.

Next, it's easiest to identify the "pi/4" angles, as they are each in the exact middle of a quadrant. Pi/4 can be marked in the middle of the first quadrant (Quadrant I), 3pi/4 is in the middle of the second quadrant, 5pi/4 is in the middle of the third quadrant, and 7pi/4 is in the middle of the fourth quadrant.

Next, the "pi/6" angles are positioned closer to the x-axis and are pi/6, 5pi/6, 7pi/6, and 11pi/6.

Finally, the "pi/3" angels are positioned closer to the y-axis and are pi/3, 2pi/3, 4pi/3, and 5pi/3.

(X,Y) POINT FOR EACH ANGLE GIVES YOU SIN AND COS trig values:

There are patterns to remember the (x,y) coordinates of the point on the trigonometric circle ("pi circle") that corresponds to each angle mentioned above.

Since each of the four "corner" points at 0, pi/2, 3pi/2, and 2pi is a distance of one full unit from the origin center of the circle, their unit circle coordinates are each (1,0), (0,1), (-1,0), and (0, -1), respectively.

NOTE: For the other angles, you only need to remember these three important numbers:

1/2

(square-root of 2)/2

(square-root of 3)/2

You just need to remember that:

1) The SMALLEST of these numbers is 1/2

2) The MID-SIZED number is (square-root of 2)/2

3) The LARGEST of these numbers is (square-root of 3)/2

For each of the remaining angles (for instance pi/6, pi/4, pi/3, etc), if the corresponding point on the circle has the smallest possible x-distance, its x-coordinate is 1/2, and if it has the largest possible x-distance, its x-coordinate is (square-root of 3)/2. If it has the middle distance, its coordinate value is (square-root of 2)/2. The same is true for the y-values.

NOTE: When doing unit circle practice, working on a unit circle worksheet, or studying for a unit circle quiz, remember that the x-coordinate is the COS value, and the y-coordinate will give you the SIN value. For instance, for the unit circle point (1,0) at angle 0, the value of cos 0 is the x-value 1, and the the value of sin 0 is the y-value 0.

Unit circle trigonometry comes up a lot in geometry, precalculus, and even calculus problems. The trig unit circle is a circle of radius one. For each angle, there is a point on the trig circle whose x-coordinate is the cosine value of the angle and whose y-coordinate is the sine value of the angle. It is a way of finding exact values of trig functions. Here is how to memorize the unit circle values, how to fill in the unit circle, and how to use the unit circle:

ANGLES (in radians):

First, the radian angle measures of the four "corner" points on the radian circle are 0, pi/2, pi, 3pi/2, and 2pi. The 2pi angle is one complete full circle around the unit circle (radians) and is in the same position as the 0 angle measure.

Next, it's easiest to identify the "pi/4" angles, as they are each in the exact middle of a quadrant. Pi/4 can be marked in the middle of the first quadrant (Quadrant I), 3pi/4 is in the middle of the second quadrant, 5pi/4 is in the middle of the third quadrant, and 7pi/4 is in the middle of the fourth quadrant.

Next, the "pi/6" angles are positioned closer to the x-axis and are pi/6, 5pi/6, 7pi/6, and 11pi/6.

Finally, the "pi/3" angels are positioned closer to the y-axis and are pi/3, 2pi/3, 4pi/3, and 5pi/3.

(X,Y) POINT FOR EACH ANGLE GIVES YOU SIN AND COS trig values:

There are patterns to remember the (x,y) coordinates of the point on the trigonometric circle ("pi circle") that corresponds to each angle mentioned above.

Since each of the four "corner" points at 0, pi/2, 3pi/2, and 2pi is a distance of one full unit from the origin center of the circle, their unit circle coordinates are each (1,0), (0,1), (-1,0), and (0, -1), respectively.

NOTE: For the other angles, you only need to remember these three important numbers:

1/2

(square-root of 2)/2

(square-root of 3)/2

You just need to remember that:

1) The SMALLEST of these numbers is 1/2

2) The MID-SIZED number is (square-root of 2)/2

3) The LARGEST of these numbers is (square-root of 3)/2

For each of the remaining angles (for instance pi/6, pi/4, pi/3, etc), if the corresponding point on the circle has the smallest possible x-distance, its x-coordinate is 1/2, and if it has the largest possible x-distance, its x-coordinate is (square-root of 3)/2. If it has the middle distance, its coordinate value is (square-root of 2)/2. The same is true for the y-values.

NOTE: When doing unit circle practice, working on a unit circle worksheet, or studying for a unit circle quiz, remember that the x-coordinate is the COS value, and the y-coordinate will give you the SIN value. For instance, for the unit circle point (1,0) at angle 0, the value of cos 0 is the x-value 1, and the the value of sin 0 is the y-value 0.

MIT grad shows how to find the inverse function of any function, if it exists. The inverse function is the reverse of your original function. It undoes whatever your function did. If your function takes x and gives you y, then the inverse function takes that y and gives you back x. Nancy formerly of MathBFF explains the steps.

How to find inverse functions: There are three steps to finding the inverse function, if it exists. The first step is to replace the f(x) with just the variable y. Second, swap the x and y variables everywhere they appear in the equation. Third, solve for y again so that you have just "y=" on one side of the equation. If this relation is a function, you can then replace the y with the "f inverse x" notation, or f^-1(x). To know if it is a function, use the Vertical Line Test or consider the form of the equation. Remember that in a function equation, for every x you input into the equation, there can only be one corresponding y value. For an example problem, jump to 00:43 in the video.

How to find inverse functions: There are three steps to finding the inverse function, if it exists. The first step is to replace the f(x) with just the variable y. Second, swap the x and y variables everywhere they appear in the equation. Third, solve for y again so that you have just "y=" on one side of the equation. If this relation is a function, you can then replace the y with the "f inverse x" notation, or f^-1(x). To know if it is a function, use the Vertical Line Test or consider the form of the equation. Remember that in a function equation, for every x you input into the equation, there can only be one corresponding y value. For an example problem, jump to 00:43 in the video.

Nancy shows how to find derivatives using the rules (Power Rule, Product Rule, Quotient Rule, etc.). To skip ahead: 1) For how and when to use the POWER RULE, constant multiple rule, constant rule, and sum and difference rule, skip to time 0:22. 2) For the PRODUCT RULE, skip to 7:36. 3) For the QUOTIENT RULE, skip to 10:53. Nancy formerly of mathbff explains the steps.

What is the derivative? It's a function that gives you the instantaneous rate of change at each point of another function. You can calculate the derivative with the definition of the derivative (using the limit), but the fastest way to find the derivative is with shortcuts such as the Power Rule, Product Rule, and Quotient Rule:

1) POWER RULE: If the given equation is a polynomial, or just a power of x, then you can use the Power Rule. For a term that's just a power of x, such as x^4, you can get the derivative by bringing down the power to the front of the term as a coefficient and decreasing the x power by 1. For example, for x^4, the derivative is 4x^3. If you have many terms added or subtracted together, and if they are powers of x, you can use the Power Rule on each term (by the Sum and Difference Rules). NOTE: The derivative of a constant, just a number, is always 0 (that is the Constant Rule). Also, if you have a term that is a constant multiplied in the front of the term, like 2x^3, you can keep the constant and differentiate the rest of the term. In this example, you keep the 2 and take the derivative of x^3, which is 3x^2, so the derivative of the term 2x^3 is 2*3x^2, or 6x^2. ANOTHER NOTE:You can use the same power rule method for fractional or negative powers, but be careful... for negative powers, it works as long as x is not 0, and for fractional/rational powers, if the power is less than 1, your derivative won't be defined at x = 0.

2) PRODUCT RULE: If your equation is not a polynomial but instead has the overall form of one expression multiplied by another expression, then you can use the Product Rule. The Product Rule says that the derivative of two functions multiplied together is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function.

3) QUOTIENT RULE: If your equation has the overall form of one expression divided by another expression, then you can use the Quotient Rule. The Quotient Rule says that the derivative of one function divided by another (a quotient) is equal to the bottom function times the derivative of the top bottom minus the top function times the derivative of the bottom function, all divided by the bottom function squared. This is true as long as the bottom function is not equal to 0.

What is the derivative? It's a function that gives you the instantaneous rate of change at each point of another function. You can calculate the derivative with the definition of the derivative (using the limit), but the fastest way to find the derivative is with shortcuts such as the Power Rule, Product Rule, and Quotient Rule:

1) POWER RULE: If the given equation is a polynomial, or just a power of x, then you can use the Power Rule. For a term that's just a power of x, such as x^4, you can get the derivative by bringing down the power to the front of the term as a coefficient and decreasing the x power by 1. For example, for x^4, the derivative is 4x^3. If you have many terms added or subtracted together, and if they are powers of x, you can use the Power Rule on each term (by the Sum and Difference Rules). NOTE: The derivative of a constant, just a number, is always 0 (that is the Constant Rule). Also, if you have a term that is a constant multiplied in the front of the term, like 2x^3, you can keep the constant and differentiate the rest of the term. In this example, you keep the 2 and take the derivative of x^3, which is 3x^2, so the derivative of the term 2x^3 is 2*3x^2, or 6x^2. ANOTHER NOTE:You can use the same power rule method for fractional or negative powers, but be careful... for negative powers, it works as long as x is not 0, and for fractional/rational powers, if the power is less than 1, your derivative won't be defined at x = 0.

2) PRODUCT RULE: If your equation is not a polynomial but instead has the overall form of one expression multiplied by another expression, then you can use the Product Rule. The Product Rule says that the derivative of two functions multiplied together is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function.

3) QUOTIENT RULE: If your equation has the overall form of one expression divided by another expression, then you can use the Quotient Rule. The Quotient Rule says that the derivative of one function divided by another (a quotient) is equal to the bottom function times the derivative of the top bottom minus the top function times the derivative of the bottom function, all divided by the bottom function squared. This is true as long as the bottom function is not equal to 0.

Nancy shows how to find any limit as x approaches a finite value/constant value (and not infinity). To skip ahead: 1) For an example of PLUGGING IN/SUBSTITUTION, skip to time 1:45. 2) For FACTORING to simplify, skip to 3:53. 3) For GETTING A COMMON DENOMINATOR, skip to time 8:09. 4) For EXPANDING by opening up parentheses to simplify and find the limit, skip to 12:01. Nancy formerly of mathbff explains the steps.

Jump to the PART 2 video (https://youtu.be/v9fQ_QeCHpI) for how to find the limit for: 5) a SQUARE ROOT in the numerator or denominator (to RATIONALIZE by multiplying by the "CONJUGATE"); 6) something of the form (SIN X)/X; or 7) an ABSOLUTE VALUE in the limit expression.

For LIMITS at INFINITY, jump to: https://youtu.be/nViVR1rImUE

1) TRY PLUGGING IN/SUBSTITUTION: The first way to try to find the limit value is to plug in for x. In the limit expression, x is approaching a certain number. If you plug in this number and get a value that is defined, then that is your limit. HOWEVER, if you get ZERO in the denominator when you plug in, then you have not found the limit yet and need to try something else to find the limit value.

2) TRY FACTORING: If you plugged in the value for x, and you got zero in the denominator (or the form 0 over 0), check whether you can factor and simplify to find the limit. If the limit expression is made up of a polynomial in a numerator and a polynomial in the denominator, then it is a very good idea to try factoring because a factor in the top may cancel with a factor in the bottom to give you a simpler expression. Then, plugging into this simpler expression may give you an actual limit value.

3) TRY GETTING A COMMON DENOMINATOR: If you plugged in the value for x, and you got zero in the denominator, and you cannot factor the expression, you have to try something else. If your limit expression has fractions within a fraction ("a complex rational expression"), try getting a common denominator in the expression. Use algebra to get a common denominator between the two fractions that are in the numerator (or denominator), and when simplifying, terms may cancel so that you have a simpler expression you can plug into to get a limit value.

4) TRY EXPANDING/OPENING UP PARENTHESES: Again, if you plugged in and got a zero in the denominator, and you can't factor or get a common denominator, consider opening up parentheses and expanding expressions by FOIL-ing or multiplying out and combining like terms. Simplifying in this way may lead to a simpler expression you can plug into to get a limit value.

Jump to the PART 2 video (https://youtu.be/v9fQ_QeCHpI) for how to find the limit for: 5) a SQUARE ROOT in the numerator or denominator (to RATIONALIZE by multiplying by the "CONJUGATE"); 6) something of the form (SIN X)/X; or 7) an ABSOLUTE VALUE in the limit expression.

For LIMITS at INFINITY, jump to: https://youtu.be/nViVR1rImUE

1) TRY PLUGGING IN/SUBSTITUTION: The first way to try to find the limit value is to plug in for x. In the limit expression, x is approaching a certain number. If you plug in this number and get a value that is defined, then that is your limit. HOWEVER, if you get ZERO in the denominator when you plug in, then you have not found the limit yet and need to try something else to find the limit value.

2) TRY FACTORING: If you plugged in the value for x, and you got zero in the denominator (or the form 0 over 0), check whether you can factor and simplify to find the limit. If the limit expression is made up of a polynomial in a numerator and a polynomial in the denominator, then it is a very good idea to try factoring because a factor in the top may cancel with a factor in the bottom to give you a simpler expression. Then, plugging into this simpler expression may give you an actual limit value.

3) TRY GETTING A COMMON DENOMINATOR: If you plugged in the value for x, and you got zero in the denominator, and you cannot factor the expression, you have to try something else. If your limit expression has fractions within a fraction ("a complex rational expression"), try getting a common denominator in the expression. Use algebra to get a common denominator between the two fractions that are in the numerator (or denominator), and when simplifying, terms may cancel so that you have a simpler expression you can plug into to get a limit value.

4) TRY EXPANDING/OPENING UP PARENTHESES: Again, if you plugged in and got a zero in the denominator, and you can't factor or get a common denominator, consider opening up parentheses and expanding expressions by FOIL-ing or multiplying out and combining like terms. Simplifying in this way may lead to a simpler expression you can plug into to get a limit value.

Nancy shows how to find the limit at a finite value with a square root, (sin x)/x, or absolute value. To skip ahead: 5) for a SQUARE ROOT in the numerator or denominator (to RATIONALIZE by multiplying by the "CONJUGATE"), skip to time 1:14. 6) for a limit with something of the form (SIN X)/X, skip to time 5:38. 7) for an ABSOLUTE VALUE in the limit expression, skip to time 14:45. Nancy formerly of mathbff explains the steps.

For PART 1 (How to Find Any Limit), jump to: https://youtu.be/nJZm-zp639s

For LIMITS at INFINITY, jump to: https://youtu.be/nViVR1rImUE

5) For a SQUARE ROOT in the numerator or denominator: If you try plugging in the value for x and get a 0 in the denominator, and you cannot factor, get a common denominator, or expand to simplify the expression, then if there's a square root in the numerator or denominator, you can try MULTIPLYING by the CONJUGATE. For instance, if you have sqrt(x+1) - 3 in the numerator, you would multiply both the numerator and denominator by sqrt(x+1) + 3 because the "conjugate" just means a two-term expression with the sign flipped in front of the second term. This is a trick or technique that helps simplify because when you multiply out, or FOIL, the numerator you will get terms that cancel. It is best to leave the denominator factored, rather than multiplying out the terms since a factor is likely to cancel. Once you simplify by multiplying on top, combining like terms, and canceling any factors from the top and bottom, try plugging in the value again for x to get an actual limit value.

6) For the form (SIN X)/X in a limit expression: If you try plugging in the value that x is approaches, and you get 0 in the denominator, if your limit expression is something of the form (sin x) over x, there is a trig property that you can use to simplify. The property is that the limit of (sin x)/x, as x approaches 0, is equal to 1. If your expression isn't exactly (sin x)/x but instead has something like 2x or 3x inside the sin function, like sin(2x) over (4x), you can use the same property but first have to rearrange the expression in a way that matches what you need, as shown in the video. NOTE: Be careful not to confuse this trig property with another, very similar, (sin x)/x expression for when x is approaching infinity. That property states that the limit of (sin x)/x, as x approaches infinity, is equal to 0. Check out the video on limits at infinity for an explanation of how to use that expression.

7) For an ABSOLUTE VALUE in your limit expression: If you try plugging in the value for x and get 0 in the denominator, and you have an absolute value in your limit expression, you will probably need to re-write the limit expression using the piecewise definition of the absolute value function. You will then have an expression for the left-side limit and one for the right-side limit. If you evaluate the left side and right side, and the numbers agree, then that is your limit value. If the two sides do not have limit values that agree, then the limit does not exist.

For PART 1 (How to Find Any Limit), jump to: https://youtu.be/nJZm-zp639s

For LIMITS at INFINITY, jump to: https://youtu.be/nViVR1rImUE

5) For a SQUARE ROOT in the numerator or denominator: If you try plugging in the value for x and get a 0 in the denominator, and you cannot factor, get a common denominator, or expand to simplify the expression, then if there's a square root in the numerator or denominator, you can try MULTIPLYING by the CONJUGATE. For instance, if you have sqrt(x+1) - 3 in the numerator, you would multiply both the numerator and denominator by sqrt(x+1) + 3 because the "conjugate" just means a two-term expression with the sign flipped in front of the second term. This is a trick or technique that helps simplify because when you multiply out, or FOIL, the numerator you will get terms that cancel. It is best to leave the denominator factored, rather than multiplying out the terms since a factor is likely to cancel. Once you simplify by multiplying on top, combining like terms, and canceling any factors from the top and bottom, try plugging in the value again for x to get an actual limit value.

6) For the form (SIN X)/X in a limit expression: If you try plugging in the value that x is approaches, and you get 0 in the denominator, if your limit expression is something of the form (sin x) over x, there is a trig property that you can use to simplify. The property is that the limit of (sin x)/x, as x approaches 0, is equal to 1. If your expression isn't exactly (sin x)/x but instead has something like 2x or 3x inside the sin function, like sin(2x) over (4x), you can use the same property but first have to rearrange the expression in a way that matches what you need, as shown in the video. NOTE: Be careful not to confuse this trig property with another, very similar, (sin x)/x expression for when x is approaching infinity. That property states that the limit of (sin x)/x, as x approaches infinity, is equal to 0. Check out the video on limits at infinity for an explanation of how to use that expression.

7) For an ABSOLUTE VALUE in your limit expression: If you try plugging in the value for x and get 0 in the denominator, and you have an absolute value in your limit expression, you will probably need to re-write the limit expression using the piecewise definition of the absolute value function. You will then have an expression for the left-side limit and one for the right-side limit. If you evaluate the left side and right side, and the numbers agree, then that is your limit value. If the two sides do not have limit values that agree, then the limit does not exist.

Nancy shows how to find the limit as x approaches infinity or negative infinity. To skip ahead: 1) For a POLYNOMIAL or CONSTANT in the limit expression, skip to 1:56. 2) For a RATIONAL ("FRACTION") expression in the limit, skip to 8:49. 3) For something of the form (SINX)/X, skip to 23:01. and 4) For an EXPONENTIAL example, skip to 27:27. Nancy formerly of mathbff explains the steps.

For LIMITS at a FINITE VALUE (not at infinity), jump to the video: https://youtu.be/nJZm-zp639s

1) For a POLYNOMIAL or CONSTANT in the limit expression: the limit of a CONSTANT (just a finite number like 3), as x approaches infinity or negative infinity, will just be equal to that same constant number. For the limit of a POLYNOMIAL (such as 2x^2 + 2x + 5), as x approaches infinity or negative infinity, just focus on the leading term (highest x power term) in the polynomial, usually the first term. You can ignore all lower terms, because as x gets infinitely large (in either the positive or negative direction), the highest term is growing most quickly, and the lower terms will not affect the limit value. Then figure out whether this leading term will grow toward positive infinity or negative infinity, as x gets extremely large. For instance, if the leading term is 2x^2, as x goes to positive infinity, this leading term will also go toward positive infinity, and the limit will be positive infinity. If the leading term were -2x^2, the x^2 would go toward infinity, as x goes to infinity, but because of the -2, the limit is negative infinity. For X approaching NEGATIVE INFINITY, keep in mind that a negative number, to an even power, becomes positive. A negative number, to an odd power, stays negative. For instance, what if the leading term is 4x^3, and you want to find the limit as x goes to negative infinity? If you think of plugging in a very large negative number for x, the 4x^3 would still be large and negative because of the odd power. The term would go toward negative infinity, so you can write that the limit is equal to negative infinity.

2) For a RATIONAL ("FRACTION") expression in the limit: I show a shortcut (and also the official formal algebraic method) to find the limit, as x goes to infinity or negative infinity. For the SHORTCUT, there are three cases: 1) If the DEGREE OF THE NUMERATOR IS LESS THAN the degree of the denominator, then the limit is equal to zero, no matter if x is approaching positive infinity or negative infinity. 2) If the DEGREE OF THE NUMERATOR IS EQUAL TO the degree of the denominator, then the limit will be equal to the ratio of the coefficients of the leading terms of the numerator an denominator, no matter if x is approaching positive infinity or negative infinity. For instance, if you're finding the limit of the rational expression (2x^2 - 5x)/(8x^2 + 3x), as x tends toward infinity or negative infinity, the limit will be equal to the ratio 2/8, which simplifies to 1/4. The limit equals 1/4. 3) If the DEGREE OF THE NUMERATOR IS GREATER THAN the degree of the denominator, then the limit will be either infinity or negative infinity. For ex., to find the limit, as x approaches infinity, of (3x^2 - 2x)/(x + 5), instead focus on finding the limit of the ratio of leading terms, as x approaches infinity. So instead, you can find the limit of 3x^2/x, which simplifies to the limit of 3x, as x approaches infinity. Since 3x goes toward infinity, as x goes to infinity, the limit is infinity. NOTE: if x had instead been approaching negative infinity, the limit of the original expression would have been negative infinity, since 3x goes to neg. infinity as x tends to neg. infinity.

3) For something of the form (SIN X)/X: there is a trig property you can use to simplify: that the limit, as x approaches infinity or neg. infinity, of (sin x)/x is equal to 0. If your expression isn't exactly (sin x)/x but instead has something like 2x or 3x inside the sine function, like sin(3x) over x, you can use the same property but first have to rearrange the expression in a way that matches what you need. Be careful not to confuse this property with another, very similar, (sin x)/x expression for when x is approaching 0. That property states that the limit of (sin x)/x, as x approaches 0, is equal to 1. Check out my video on how to find the limit, at a finite value, for how to use that property.

4) For an EXPONENTIAL in your limit expression (with a negative power): for instance, if you are finding the limit, as x approaches infinity, of e^(-2x), first rewrite the expression using the reciprocal instead of the negative power, so 1/e^(2x). Then it is easier to see what happens as x gets extremely large and goes toward infinity. The e^(2x) gets extremely large, so 1 over a very large number will head toward zero, and the limit will be equal to 0.

For LIMITS at a FINITE VALUE (not at infinity), jump to the video: https://youtu.be/nJZm-zp639s

1) For a POLYNOMIAL or CONSTANT in the limit expression: the limit of a CONSTANT (just a finite number like 3), as x approaches infinity or negative infinity, will just be equal to that same constant number. For the limit of a POLYNOMIAL (such as 2x^2 + 2x + 5), as x approaches infinity or negative infinity, just focus on the leading term (highest x power term) in the polynomial, usually the first term. You can ignore all lower terms, because as x gets infinitely large (in either the positive or negative direction), the highest term is growing most quickly, and the lower terms will not affect the limit value. Then figure out whether this leading term will grow toward positive infinity or negative infinity, as x gets extremely large. For instance, if the leading term is 2x^2, as x goes to positive infinity, this leading term will also go toward positive infinity, and the limit will be positive infinity. If the leading term were -2x^2, the x^2 would go toward infinity, as x goes to infinity, but because of the -2, the limit is negative infinity. For X approaching NEGATIVE INFINITY, keep in mind that a negative number, to an even power, becomes positive. A negative number, to an odd power, stays negative. For instance, what if the leading term is 4x^3, and you want to find the limit as x goes to negative infinity? If you think of plugging in a very large negative number for x, the 4x^3 would still be large and negative because of the odd power. The term would go toward negative infinity, so you can write that the limit is equal to negative infinity.

2) For a RATIONAL ("FRACTION") expression in the limit: I show a shortcut (and also the official formal algebraic method) to find the limit, as x goes to infinity or negative infinity. For the SHORTCUT, there are three cases: 1) If the DEGREE OF THE NUMERATOR IS LESS THAN the degree of the denominator, then the limit is equal to zero, no matter if x is approaching positive infinity or negative infinity. 2) If the DEGREE OF THE NUMERATOR IS EQUAL TO the degree of the denominator, then the limit will be equal to the ratio of the coefficients of the leading terms of the numerator an denominator, no matter if x is approaching positive infinity or negative infinity. For instance, if you're finding the limit of the rational expression (2x^2 - 5x)/(8x^2 + 3x), as x tends toward infinity or negative infinity, the limit will be equal to the ratio 2/8, which simplifies to 1/4. The limit equals 1/4. 3) If the DEGREE OF THE NUMERATOR IS GREATER THAN the degree of the denominator, then the limit will be either infinity or negative infinity. For ex., to find the limit, as x approaches infinity, of (3x^2 - 2x)/(x + 5), instead focus on finding the limit of the ratio of leading terms, as x approaches infinity. So instead, you can find the limit of 3x^2/x, which simplifies to the limit of 3x, as x approaches infinity. Since 3x goes toward infinity, as x goes to infinity, the limit is infinity. NOTE: if x had instead been approaching negative infinity, the limit of the original expression would have been negative infinity, since 3x goes to neg. infinity as x tends to neg. infinity.

3) For something of the form (SIN X)/X: there is a trig property you can use to simplify: that the limit, as x approaches infinity or neg. infinity, of (sin x)/x is equal to 0. If your expression isn't exactly (sin x)/x but instead has something like 2x or 3x inside the sine function, like sin(3x) over x, you can use the same property but first have to rearrange the expression in a way that matches what you need. Be careful not to confuse this property with another, very similar, (sin x)/x expression for when x is approaching 0. That property states that the limit of (sin x)/x, as x approaches 0, is equal to 1. Check out my video on how to find the limit, at a finite value, for how to use that property.

4) For an EXPONENTIAL in your limit expression (with a negative power): for instance, if you are finding the limit, as x approaches infinity, of e^(-2x), first rewrite the expression using the reciprocal instead of the negative power, so 1/e^(2x). Then it is easier to see what happens as x gets extremely large and goes toward infinity. The e^(2x) gets extremely large, so 1 over a very large number will head toward zero, and the limit will be equal to 0.

Nancy shows how to find the equation of a tangent line using derivatives (Calculus). To skip ahead: 1) For a BASIC example, skip to time 0:44. 2) For a more complex example that uses the CHAIN RULE, skip to time 5:58. 3) For an example that uses TRIG FUNCTIONS, skip to time 11:35. Nancy formerly of mathbff explains the steps.

For HOW TO FIND DERIVATIVES, jump to: https://youtu.be/QqF3i1pnyzU

What is a tangent line? It is a line that is tangent to a curve at one point. You can use calculus to find the tangent line by taking the derivative of the given equation. Here are the steps to finding the equation of the tangent line, using derivatives:

1) TAKE DERIVATIVE: The first step is to take the derivative of the given equation, with respect to x. For instance, to find the equation of the line tangent to f(x) = x^2 + 3 at x = 4, first take the derivative of f(x), which is 2x.

2) PLUG IN X-VALUE INTO DERIVATIVE TO GET SLOPE: The second step is to plug the given x-value into the derivative of f(x). The value you get is the slope of the tangent line, m.

3) FIND Y-VALUE OF POINT WITH ORIGINAL EQUATION: The third step is to find the y-value of the point. You get this by plugging the given x-value into the original equation to find the corresponding y-value. Since you want the tangent line at x = 4, plug x = 4 into the original f(x), and you get a y-value of 7. This means that the full point is (4,7).

4) PLUG THE SLOPE AND X,Y POINT VALUES INTO POINT-SLOPE EQUATION: The last step is to put the x and y point values, as well as the slope, m, into the point-slope equation of a line. The point-slope equation is x - x1 = m (y - y1). Plug the x and y point values into this equation for x1 and y1. Plug the slope you found in for m. This is the equation of the tangent line. The equation will still have x and y variables in it.

You can also rearrange the equation you get so that it is in "y equals" or slope-intercept form, if you want.

For HOW TO FIND DERIVATIVES, jump to: https://youtu.be/QqF3i1pnyzU

What is a tangent line? It is a line that is tangent to a curve at one point. You can use calculus to find the tangent line by taking the derivative of the given equation. Here are the steps to finding the equation of the tangent line, using derivatives:

1) TAKE DERIVATIVE: The first step is to take the derivative of the given equation, with respect to x. For instance, to find the equation of the line tangent to f(x) = x^2 + 3 at x = 4, first take the derivative of f(x), which is 2x.

2) PLUG IN X-VALUE INTO DERIVATIVE TO GET SLOPE: The second step is to plug the given x-value into the derivative of f(x). The value you get is the slope of the tangent line, m.

3) FIND Y-VALUE OF POINT WITH ORIGINAL EQUATION: The third step is to find the y-value of the point. You get this by plugging the given x-value into the original equation to find the corresponding y-value. Since you want the tangent line at x = 4, plug x = 4 into the original f(x), and you get a y-value of 7. This means that the full point is (4,7).

4) PLUG THE SLOPE AND X,Y POINT VALUES INTO POINT-SLOPE EQUATION: The last step is to put the x and y point values, as well as the slope, m, into the point-slope equation of a line. The point-slope equation is x - x1 = m (y - y1). Plug the x and y point values into this equation for x1 and y1. Plug the slope you found in for m. This is the equation of the tangent line. The equation will still have x and y variables in it.

You can also rearrange the equation you get so that it is in "y equals" or slope-intercept form, if you want.

Nancy shows how to use the substitution method to solve a system of linear equations (aka. simultaneous equations). To skip ahead: 1) For a BASIC SUBSTITUTION example, skip to time 0:19. 2) For an example where it's TRICKIER to solve and also involves FRACTIONS, skip to 08:17. and P.S.) For WEIRDER-LOOKING DIFFERENT FORMS of systems, for which you can still use substitution, skip to 16:46. Nancy formerly of mathbff explains the steps.

For how to solve a system of linear equations with the ELIMINATION METHOD, jump to: https://youtu.be/XOJgzW4P7T8

What does it mean to solve a system of equations using the SUBSTITUTION METHOD? It means to solve for the x and y values that make both equations true, using the substitution method. Substitution is just one way of solving (elimination is another way). Here are the THREE STEPS you can always follow:

STEP 1) First, solve for x or y in either of the two equations in the system (get x or y alone on one side). It doesn't matter which equation you use first, and it doesn't matter whether you get x alone first or y alone first. You can pick whatever looks easiest to solve for in the system.

STEP 2) Then, take the expression you found for y (or x) and plug it into the other equation you haven't used yet. For instance, if you got y alone in the first step, take that expression for y and plug it into, or substitute it into, the variable y in the other original equation in the system. Then, simplify the expression by distributing to open up the parentheses (using the distributive property) and combining like terms. Solve for the unknown variable in this equation.

STEP 3) Finally, once you solve and get an actual number for a variable, you can take that number and plug it back into either original equation in the system and then simplify and solve for the other variable. This (x,y) pair is your solution.

IMPORTANT: If you get something like 0 = 0 or 8 = 8, or something like 1 = 2 or 0 = 21 when you are following these steps and trying to solve, then you've probably found a special case. If you get something TRUE like 0 = 0 (a number equals the same number), then there are an infinite number of solutions to the system, and you can just write "infinitely-many solutions". If instead you get something FALSE like 0 = 21, then there is no solution (inconsistent system), and you can write "no solution".

For how to solve a system of linear equations with the ELIMINATION METHOD, jump to: https://youtu.be/XOJgzW4P7T8

What does it mean to solve a system of equations using the SUBSTITUTION METHOD? It means to solve for the x and y values that make both equations true, using the substitution method. Substitution is just one way of solving (elimination is another way). Here are the THREE STEPS you can always follow:

STEP 1) First, solve for x or y in either of the two equations in the system (get x or y alone on one side). It doesn't matter which equation you use first, and it doesn't matter whether you get x alone first or y alone first. You can pick whatever looks easiest to solve for in the system.

STEP 2) Then, take the expression you found for y (or x) and plug it into the other equation you haven't used yet. For instance, if you got y alone in the first step, take that expression for y and plug it into, or substitute it into, the variable y in the other original equation in the system. Then, simplify the expression by distributing to open up the parentheses (using the distributive property) and combining like terms. Solve for the unknown variable in this equation.

STEP 3) Finally, once you solve and get an actual number for a variable, you can take that number and plug it back into either original equation in the system and then simplify and solve for the other variable. This (x,y) pair is your solution.

IMPORTANT: If you get something like 0 = 0 or 8 = 8, or something like 1 = 2 or 0 = 21 when you are following these steps and trying to solve, then you've probably found a special case. If you get something TRUE like 0 = 0 (a number equals the same number), then there are an infinite number of solutions to the system, and you can just write "infinitely-many solutions". If instead you get something FALSE like 0 = 21, then there is no solution (inconsistent system), and you can write "no solution".

Nancy shows how to use the elimination method to solve a system of linear equations (aka. simultaneous equations). To skip ahead: 1) For a BASIC example where terms cancel right away when you add the equations, skip to 0:25. 2) For an example in which you have to MULTIPLY ONE EQUATION by a number before adding the equations, skip to time 6:26. 3) For an example in which you have to MULTIPLY BOTH EQUATIONS by numbers before adding, skip to 12:12. P.S.) For HOW TO SUBTRACT equations instead of adding them, if you'd rather do that, skip to 18:00. Nancy formerly of mathbff explains the steps.

For how to solve a system of linear equations by the SUBSTITUTION METHOD, jump to: https://youtu.be/YriMMWbndn0

What does it mean to solve a system of equations using the elimination method? It means to solve for the x and y values that make both equations true, using the elimination method. Elimination is just one way of solving (substitution is another way) and is sometimes called linear combination. Here's how:

1) BASIC ELIMINATION: First, make sure the equations are ordered in the same way, with their like terms lined up vertically. Then, try adding the equations (vertically) to make a new equation. If a term in the first equation has the same number coefficient but opposite sign (like 5y and -5y), then adding the equations will mean that those terms cancel. Since those terms disappear (in the example, 5y and -5y), there is just the variable x left in the equation so you can solve for x. Once you solve for x, you can take that number and plug it in for x in either of the original equations in the system, to solve for the other variable, y. The (x,y) pair is your solution.

2) MULTIPLY ONE EQUATION BEFORE ADDING: Sometimes terms don't immediately cancel when you add the two equations. What do you do then? Try multiplying one of the equations by a number (both the left and right sides) to make either the x terms or y terms have similar coefficients (but opposite sign), like positive 4x and negative 4x. Then add the two equations (the new version) and follow the same steps as in the first example.

3) MULTIPLY BOTH EQUATIONS BEFORE ADDING: If it does not help to just multiply one equation by a number, you may need to multiply both equations, separately, by numbers. If you check to see whether you can just multiply one equation by a number and add opposite terms/eliminate a variable, and it is not possible, then aim for an LCM, or least common multiple, by multiplying each equation by a different number, and then adding the equations.

P.S.) YES, YOU CAN SUBTRACT INSTEAD (if you want): If you have a system that has the exact same x terms or exact same y terms in both equations, if you prefer, you can just subtract the equations. Alternatively, if you still wanted to add the equations, you could just multiply one of the equations by -1 first and then add the equations and solve normally.

IMPORTANT: If you get something like 0 = 0 or 8 = 8, or something like 1 = 2 or 0 = 21 when you are following these steps and trying to solve, then you have probably found a special case. If you get something TRUE like 0 = 0 (a number equals the same number), then there are an infinite number of solutions to the system, and you can just write "infinitely-many solutions". If instead you get something FALSE like 0 = 21, then there is no solution (inconsistent system), and you can write "no solution".

For how to solve a system of linear equations by the SUBSTITUTION METHOD, jump to: https://youtu.be/YriMMWbndn0

What does it mean to solve a system of equations using the elimination method? It means to solve for the x and y values that make both equations true, using the elimination method. Elimination is just one way of solving (substitution is another way) and is sometimes called linear combination. Here's how:

1) BASIC ELIMINATION: First, make sure the equations are ordered in the same way, with their like terms lined up vertically. Then, try adding the equations (vertically) to make a new equation. If a term in the first equation has the same number coefficient but opposite sign (like 5y and -5y), then adding the equations will mean that those terms cancel. Since those terms disappear (in the example, 5y and -5y), there is just the variable x left in the equation so you can solve for x. Once you solve for x, you can take that number and plug it in for x in either of the original equations in the system, to solve for the other variable, y. The (x,y) pair is your solution.

2) MULTIPLY ONE EQUATION BEFORE ADDING: Sometimes terms don't immediately cancel when you add the two equations. What do you do then? Try multiplying one of the equations by a number (both the left and right sides) to make either the x terms or y terms have similar coefficients (but opposite sign), like positive 4x and negative 4x. Then add the two equations (the new version) and follow the same steps as in the first example.

3) MULTIPLY BOTH EQUATIONS BEFORE ADDING: If it does not help to just multiply one equation by a number, you may need to multiply both equations, separately, by numbers. If you check to see whether you can just multiply one equation by a number and add opposite terms/eliminate a variable, and it is not possible, then aim for an LCM, or least common multiple, by multiplying each equation by a different number, and then adding the equations.

P.S.) YES, YOU CAN SUBTRACT INSTEAD (if you want): If you have a system that has the exact same x terms or exact same y terms in both equations, if you prefer, you can just subtract the equations. Alternatively, if you still wanted to add the equations, you could just multiply one of the equations by -1 first and then add the equations and solve normally.

IMPORTANT: If you get something like 0 = 0 or 8 = 8, or something like 1 = 2 or 0 = 21 when you are following these steps and trying to solve, then you have probably found a special case. If you get something TRUE like 0 = 0 (a number equals the same number), then there are an infinite number of solutions to the system, and you can just write "infinitely-many solutions". If instead you get something FALSE like 0 = 21, then there is no solution (inconsistent system), and you can write "no solution".

Nancy shows how to find the vertical asymptotes of a rational function and what they look like on a graph. To skip ahead: 1) For the STEPS TO FIND THE VERTICAL ASYMPTOTE(S) and an example with two vertical asymptotes, skip to 0:19. 2) For an example in which FACTORS CANCEL and that has one vertical asymptote and a HOLE, skip to 5:58. 3) For an example with NO VERTICAL ASYMPTOTES, skip to time 10:12. Nancy formerly of mathbff explains the steps.

For how to find the HORIZONTAL ASYMPTOTE jump to: https://youtu.be/qJrrZQgSkO8

For how to FACTOR quadratics jump to: https://youtu.be/YtN9_tCaRQc

For how to find the DOMAIN of a function jump to: https://youtu.be/GQGFMUfr10M

What is a vertical asymptote? It's an invisible vertical line that a function gets really really close to but never reaches. How do you find the vertical asymptote(s) from the given equation?

THREE STEPS TO FIND THE VERTICAL ASYMPTOTE(S): For a rational function, there are three main steps you can always follow to find all the vertical asymptotes, if there are any:

STEP 1) FACTOR: The first step is to factor the top and bottom (numerator and denominator) if you can, and as much as you can. For instance, in the function f(x) = (x^2 + 3x - 10)/(x^2 - 4), you can factor both the top and bottom. The numerator, x^2 + 3x - 10, is a quadratic that factors into (x + 5)(x - 2), and the denominator, x^2 - 4, is a difference of squares that factors into (x + 2)(x - 2). You then rewrite the whole function with both of these factorizations so that you have f(x) = [(x + 5)(x - 2)] / [(x + 2)(x - 2)].

STEP 2) CANCEL: Next, simplify the function by canceling any factors that are the same on top and bottom. If there are no common factors, you can leave it alone. In our example from Step 1, there is an x - 2 term on both the top and bottom, so we can cancel those two factors. You can rewrite the function after getting rid of those similar factors so that it looks like: f(x) = (x + 5)/(x + 2).

STEP 3) SET THE DENOMINATOR EQUAL TO ZERO: After simplifying and getting rid of any common factors, the last step is to find the real zeros of the denominator by taking the bottom of the simplified function and setting it equal to zero. You then solve that equation for x, and any real numbers you get as a solution for x are where there are vertical asymptotes. You can write your answers as just "x equals [some number]". If you have vertical asymptotes, they will always be in that form, such as x = 3 or x = -2. These represent vertical (invisible) lines on the graph that your function approaches but never crosses.

Remember that if you get an imaginary answer when you solve for x (such as a square root of a negative number), then there are no vertical asymptotes. If there is no real solution when you solve for x, then there are NO VERTICAL ASYMPTOTES. Note: By the way, if you had factors that cancelled in Step 2, that created a "hole", or removable discontinuity, on the graph where the function was indeterminate.

For how to find the HORIZONTAL ASYMPTOTE jump to: https://youtu.be/qJrrZQgSkO8

For how to FACTOR quadratics jump to: https://youtu.be/YtN9_tCaRQc

For how to find the DOMAIN of a function jump to: https://youtu.be/GQGFMUfr10M

What is a vertical asymptote? It's an invisible vertical line that a function gets really really close to but never reaches. How do you find the vertical asymptote(s) from the given equation?

THREE STEPS TO FIND THE VERTICAL ASYMPTOTE(S): For a rational function, there are three main steps you can always follow to find all the vertical asymptotes, if there are any:

STEP 1) FACTOR: The first step is to factor the top and bottom (numerator and denominator) if you can, and as much as you can. For instance, in the function f(x) = (x^2 + 3x - 10)/(x^2 - 4), you can factor both the top and bottom. The numerator, x^2 + 3x - 10, is a quadratic that factors into (x + 5)(x - 2), and the denominator, x^2 - 4, is a difference of squares that factors into (x + 2)(x - 2). You then rewrite the whole function with both of these factorizations so that you have f(x) = [(x + 5)(x - 2)] / [(x + 2)(x - 2)].

STEP 2) CANCEL: Next, simplify the function by canceling any factors that are the same on top and bottom. If there are no common factors, you can leave it alone. In our example from Step 1, there is an x - 2 term on both the top and bottom, so we can cancel those two factors. You can rewrite the function after getting rid of those similar factors so that it looks like: f(x) = (x + 5)/(x + 2).

STEP 3) SET THE DENOMINATOR EQUAL TO ZERO: After simplifying and getting rid of any common factors, the last step is to find the real zeros of the denominator by taking the bottom of the simplified function and setting it equal to zero. You then solve that equation for x, and any real numbers you get as a solution for x are where there are vertical asymptotes. You can write your answers as just "x equals [some number]". If you have vertical asymptotes, they will always be in that form, such as x = 3 or x = -2. These represent vertical (invisible) lines on the graph that your function approaches but never crosses.

Remember that if you get an imaginary answer when you solve for x (such as a square root of a negative number), then there are no vertical asymptotes. If there is no real solution when you solve for x, then there are NO VERTICAL ASYMPTOTES. Note: By the way, if you had factors that cancelled in Step 2, that created a "hole", or removable discontinuity, on the graph where the function was indeterminate.

Nancy shows how to find sin, cos, and tan using SohCahToa as well as the csc, sec, and cot trig functions. Nancy formerly of mathbff explains the steps.

For MORE TRIG: how to convert between radians and degrees, jump to: https://youtu.be/l6hSY2Pcch0

If you have a right triangle with side lengths given and an angle (theta), you can easily find the sine, cosine, and tangent trig functions. Remember that the longest side of the triangle is the 'HYPOTENUSE', the side directly opposite theta is the 'OPPOSITE', and the other side next to theta is the 'ADJACENT' side.

Once you've identified the hypotenuse (H), adjacent (A), and opposite sides (O), you can find Sin, Cos, and Tan:

Sin of theta = Opposite over Hypotenuse (SOH)

Cos of theta = Adjacent over Hypotenuse (CAH)

Tan of theta = Opposite over Adjacent (TOA)

The mnemonic SOH CAH TOA can help you remember which sides to use for sin, cos, and tan. For instance, SOH stands for Sin = Opposite over Hypotenuse

If you need to find one of the other three trig functions (Csc, Sec, or Cot), the easiest way to find them is to first find Sin, Cos, or Tan and then take the reciprocal (or flip) the fraction, since:

Csc = 1 / Sin

Sec = 1 / Cos

Cot = 1 / Tan

For example, if you find that Cos of an angle is equal to 3/5, the Sec of that angle would be equal to 5/3.

For MORE TRIG: how to convert between radians and degrees, jump to: https://youtu.be/l6hSY2Pcch0

If you have a right triangle with side lengths given and an angle (theta), you can easily find the sine, cosine, and tangent trig functions. Remember that the longest side of the triangle is the 'HYPOTENUSE', the side directly opposite theta is the 'OPPOSITE', and the other side next to theta is the 'ADJACENT' side.

Once you've identified the hypotenuse (H), adjacent (A), and opposite sides (O), you can find Sin, Cos, and Tan:

Sin of theta = Opposite over Hypotenuse (SOH)

Cos of theta = Adjacent over Hypotenuse (CAH)

Tan of theta = Opposite over Adjacent (TOA)

The mnemonic SOH CAH TOA can help you remember which sides to use for sin, cos, and tan. For instance, SOH stands for Sin = Opposite over Hypotenuse

If you need to find one of the other three trig functions (Csc, Sec, or Cot), the easiest way to find them is to first find Sin, Cos, or Tan and then take the reciprocal (or flip) the fraction, since:

Csc = 1 / Sin

Sec = 1 / Cos

Cot = 1 / Tan

For example, if you find that Cos of an angle is equal to 3/5, the Sec of that angle would be equal to 5/3.

MIT grad shows how to convert from radians to degrees, convert from degrees to radians, and explains what radians and degrees mean. To skip ahead: 1) For converting from DEGREES TO RADIANS, skip to time 4:50. 2) For converting from RADIANS TO DEGREES, skip to time 8:11. 3) For CONVERTING A DECIMAL number of radians into degrees, skip to time 10:50. Nancy formerly of mathbff explains the steps.

For help with basic TRIGONOMETRY (SIN, COS, TAN), jump to: https://youtu.be/bSM7RNSbWhM

Radians and degrees are just two different ways of expressing an angle, and they have different units. Angle measures written in degrees have a little superscript circle symbol after the number. Angle measures written in radians do not have the degree circle symbol and may or may not have the units "radians" or the abbreviation "rad." listed after the number.

1) TO CONVERT FROM DEGREES TO RADIANS: Multiply the number of degrees by the conversion factor PI OVER 180. For instance, to convert from 200 degrees into radians, you would multiply 200 degrees by the fraction quantity pi radians over 180 degrees. Since you are starting with degrees, and your conversion factor has degrees in the denominator, the degrees units will go away, or effectively cancel, and you will be left with the radian units you want. 200 times pi, divided by 180, simplifies to the quantity 10 pi over 9, which is now in radians. You can leave the symbol pi in your final answer.

2) TO CONVERT FROM RADIANS TO DEGREES: Multiply the number of radians by the conversion factor 180 OVER PI. For instance, to convert from 2 pi over 9 radians into degrees, you would multiply 2 pi over 9 by the fraction quantity 180 degrees over pi radians. Since you are starting with degrees, and your conversion factor has radians in the denominator, the radians units go away, or cancel out, and you will be left with the degree units you want. Since you are also starting with a pi in your angle measure, and the conversion factor has a pi in the denominator, the pi symbols will cancel out and you're left with just 2 times 180 over 9. 2 times 180 is 360, and 360 divided by 9 simplifies to 40, so your final answer is 40 degrees.

What is a radian? Technically, the definition is that if you had a circle with a radius r, an arc length distance of r around the circle would create an inside, central angle that is one radian.

How many degrees are in a circle? There are 360 degrees total in one full circle. Starting from zero degrees at the right-most point on the circle and moving counter-clockwise, one quarter of the way through a circle is 90 degrees, one half of the circle is 180 degrees, three-quarters of the way is 270 degrees, and one full circle is 360 degrees and brings you all the way back to your starting 0 degree point.

How many radians are in a circle? There are 2 pi radians total in one full circle. Starting from zero radians at the right-most point on the circle moving counter-clockwise, one quarter of the way through a circle is pi over 2 radians, one half of the circle is pi radians, three-quarters of the way is 3 pi over 2 radians, and one full circle is 2 pi radians and brings you back to your starting 0 radian point.

Notice that 360 degrees is equivalent to, or the same as, 2 pi. Since 360 degrees equals 2 pi, you could divide both sides of this equation by 2 and find that 180 degrees equals pi. You will use this equivalence, the fact that 180 degrees equals pi, to convert from degrees to radians, or from radians to degrees.

For help with basic TRIGONOMETRY (SIN, COS, TAN), jump to: https://youtu.be/bSM7RNSbWhM

Radians and degrees are just two different ways of expressing an angle, and they have different units. Angle measures written in degrees have a little superscript circle symbol after the number. Angle measures written in radians do not have the degree circle symbol and may or may not have the units "radians" or the abbreviation "rad." listed after the number.

1) TO CONVERT FROM DEGREES TO RADIANS: Multiply the number of degrees by the conversion factor PI OVER 180. For instance, to convert from 200 degrees into radians, you would multiply 200 degrees by the fraction quantity pi radians over 180 degrees. Since you are starting with degrees, and your conversion factor has degrees in the denominator, the degrees units will go away, or effectively cancel, and you will be left with the radian units you want. 200 times pi, divided by 180, simplifies to the quantity 10 pi over 9, which is now in radians. You can leave the symbol pi in your final answer.

2) TO CONVERT FROM RADIANS TO DEGREES: Multiply the number of radians by the conversion factor 180 OVER PI. For instance, to convert from 2 pi over 9 radians into degrees, you would multiply 2 pi over 9 by the fraction quantity 180 degrees over pi radians. Since you are starting with degrees, and your conversion factor has radians in the denominator, the radians units go away, or cancel out, and you will be left with the degree units you want. Since you are also starting with a pi in your angle measure, and the conversion factor has a pi in the denominator, the pi symbols will cancel out and you're left with just 2 times 180 over 9. 2 times 180 is 360, and 360 divided by 9 simplifies to 40, so your final answer is 40 degrees.

What is a radian? Technically, the definition is that if you had a circle with a radius r, an arc length distance of r around the circle would create an inside, central angle that is one radian.

How many degrees are in a circle? There are 360 degrees total in one full circle. Starting from zero degrees at the right-most point on the circle and moving counter-clockwise, one quarter of the way through a circle is 90 degrees, one half of the circle is 180 degrees, three-quarters of the way is 270 degrees, and one full circle is 360 degrees and brings you all the way back to your starting 0 degree point.

How many radians are in a circle? There are 2 pi radians total in one full circle. Starting from zero radians at the right-most point on the circle moving counter-clockwise, one quarter of the way through a circle is pi over 2 radians, one half of the circle is pi radians, three-quarters of the way is 3 pi over 2 radians, and one full circle is 2 pi radians and brings you back to your starting 0 radian point.

Notice that 360 degrees is equivalent to, or the same as, 2 pi. Since 360 degrees equals 2 pi, you could divide both sides of this equation by 2 and find that 180 degrees equals pi. You will use this equivalence, the fact that 180 degrees equals pi, to convert from degrees to radians, or from radians to degrees.

Nancy shows how to simplify radical expressions, specifically square root expressions, into their simplest form ("Simplified Radical Form" or "SRF Form"). To skip ahead: 1) for a PERFECT SQUARE under the root like sqrt(16) skip to time 1:29. 2) for a SMALL number under the root that is NOT a perfect square like sqrt(32), skip to time 2:45. 3) for a LARGE number under the root, such as sqrt(343), skip to 5:14. 4) for a PRODUCT, meaning a number times a square root like 5*sqrt(54) skip to 7:30. 5) for a QUOTIENT, meaning a number divided by a square root like 4/sqrt(2), and where you have to RATIONALIZE the denominator, skip to 10:10. and 6) for a more complex QUOTIENT example like (4+sqrt(3))/(5-sqrt(3)) where you have to multiply by the CONJUGATE to simplify, skip to 12:36. Nancy formerly of mathbff explains the steps.

For all simplifying radical problems:

1) First try to find a perfect square number that divides evenly into the number under the square root. Perfect square numbers include 4, 9, 16, 25, 36, 49, etc. because they are squares of the numbers 2, 3, 4, 5, 6, 7, etc., respectively.

2) Separate the root expression into the two numbers you found as factors, with each number now under its own square root symbol. For example, sqrt(12) = sqrt(4) times sqrt(3).

3) Keep simplifying. The square root of a perfect square becomes just a number, without the radical square root symbol. sqrt(4)*sqrt(3) becomes 2*sqrt(3).

4) Check to see if anymore simplifying can be done, either by combining constants or finding another perfect square number that divides evenly into your factors. If no more simplifying can be done, you have your final "simplified radical form" (SRF) answer, 2*sqrt(3).

If there is a square root anywhere in the DENOMINATOR, you must "rationalize" the denominator, in other words, somehow get rid of the root in the bottom, in order for the expression to be considered simplest. For example, if you have 4/sqrt(2) you must use the trick of simplifying by multiplying the expression by sqrt(2)/sqrt(2). This will clear the square root in the bottom. If you multiply straight across, you would get 4*sqrt(2)/2, which simplifies again to 2*sqrt(2).

If you have a more complicated fraction with a square root in the denominator, like (4+sqrt(3))/(5-sqrt(3)), you will need to use the "conjugate" to simplify. This means multiplying top and bottom by the expression in the denominator, but with the middle sign flipped. Here you would multiply by (5+sqrt(3))/(5+sqrt(3)), multiply straight across on top and bottom using FOIL-ing and distribution. Some terms will cancel in the bottom, leaving you with a simpler denominator that has no square roots.

For all simplifying radical problems:

1) First try to find a perfect square number that divides evenly into the number under the square root. Perfect square numbers include 4, 9, 16, 25, 36, 49, etc. because they are squares of the numbers 2, 3, 4, 5, 6, 7, etc., respectively.

2) Separate the root expression into the two numbers you found as factors, with each number now under its own square root symbol. For example, sqrt(12) = sqrt(4) times sqrt(3).

3) Keep simplifying. The square root of a perfect square becomes just a number, without the radical square root symbol. sqrt(4)*sqrt(3) becomes 2*sqrt(3).

4) Check to see if anymore simplifying can be done, either by combining constants or finding another perfect square number that divides evenly into your factors. If no more simplifying can be done, you have your final "simplified radical form" (SRF) answer, 2*sqrt(3).

If there is a square root anywhere in the DENOMINATOR, you must "rationalize" the denominator, in other words, somehow get rid of the root in the bottom, in order for the expression to be considered simplest. For example, if you have 4/sqrt(2) you must use the trick of simplifying by multiplying the expression by sqrt(2)/sqrt(2). This will clear the square root in the bottom. If you multiply straight across, you would get 4*sqrt(2)/2, which simplifies again to 2*sqrt(2).

If you have a more complicated fraction with a square root in the denominator, like (4+sqrt(3))/(5-sqrt(3)), you will need to use the "conjugate" to simplify. This means multiplying top and bottom by the expression in the denominator, but with the middle sign flipped. Here you would multiply by (5+sqrt(3))/(5+sqrt(3)), multiply straight across on top and bottom using FOIL-ing and distribution. Some terms will cancel in the bottom, leaving you with a simpler denominator that has no square roots.

MIT grad explains solving inequalities. This video focuses on solving linear inequalities. It shows when to switch the sign of the inequality, if you divide or multiply by a negative number, and is an introduction to how to solve inequalities in algebra. To skip ahead: 1) For a basic example of SOLVING AN INEQUALITY, skip to time 00:34. 2) For when you have to FLIP THE INEQUALITY SIGN when DIVIDING BY A NEGATIVE number, skip to 1:26. 3) For when you need to reverse the inequality symbol because you're MULTIPLYING BY A NEGATIVE number, skip to 2:42. The problems here are multi-step inequalities, or two-step inequalities. Nancy formerly of MathBFF explains the steps.

Algebraic equations and inequalities: what is inequality math? What does inequality mean? An inequality is a number sentence that uses an inequality symbol instead of an equality symbol (=), which means it has a greater than sign, less than sign, greater than or equal to sign, or less than or equal to sign, instead of the normal equal sign in an equation. One linear inequality definition is: an inequality that involves a linear algebraic expression and inequality signs.

How to find the solution of an inequality algebra problem:

1) A simple, BASIC LINEAR INEQUALITY example you might see on a solving inequalities worksheet or problem set is the expression 2x + 3 [less than or equal to] 11. You can solve for x just like you would for an equality, or normal equation, by trying to get x alone. This is an example of two-step inequalities, or multi-step inequalities. The first step is to subtract the constant 3 from both sides, in order to move it to the right hand side, so that you have 2x [less than or equal to] 8. The second step is to get x alone on the left side by dividing both sides by 2 so that your answer for the inequality is x [less than or equal to] 4. In this case, solving the inequality used the exact same steps you would use to solve an "equal sign" equation.

Sometimes the steps are different from solving normal math equations. Solving for x is the same as solving a linear equality (-2x + 3 = 11) except that if you ever have to divide by a negative number or multiply by a negative number while solving for x, you must flip the sign, or switch the direction of your inequality symbol less than becomes greater than, for example). Here are the two cases of inequality problems WHEN YOU NEED TO REVERSE THE SIGN of the inequality because you're multiplying or dividing both sides of the inequality by a negative number:

2) DIVIDING EXAMPLE: For an inequality like -2x + 3 [less than or equal to] 11, after moving the constant 3 to the right side, you will need to divide both sides by negative 2 in order to get x alone on the left side for the solution. WHEN YOU DIVIDE BY A NEGATIVE NUMBER, you need to reverse the direction of the inequality symbol in your expression. In this example, after you divide out -2, you get x [greater than or equal to] -4 (instead of x less than or equal to -4) for the final solution. This example was also a two-step inequality problem.

3) MULTIPLYING EXAMPLE: For examples like (-1/2)x + 4 [greater than] 7, when you've moved the constant to the right side and have the inequality (-1/2)x [greater than] 3, the fastest way to get x alone is to clear the fraction by multiplying both sides by -2. This will cancel the 2 in the denominator and remove the negative sign in front of x on the left side. But WHEN YOU MULTIPLY BY A NEGATIVE, you have to flip the direction of the inequality symbol so that your answer is x [less than] -6 instead of x [greater than] -6. This problem was also a multi-step inequality.

Note on inequality rules: you don't have to switch the symbol if you're adding a negative number or subtracting a negative number: switching the inequality sign is only for when you're multiplying or dividing by a negative number.

Algebraic equations and inequalities: what is inequality math? What does inequality mean? An inequality is a number sentence that uses an inequality symbol instead of an equality symbol (=), which means it has a greater than sign, less than sign, greater than or equal to sign, or less than or equal to sign, instead of the normal equal sign in an equation. One linear inequality definition is: an inequality that involves a linear algebraic expression and inequality signs.

How to find the solution of an inequality algebra problem:

1) A simple, BASIC LINEAR INEQUALITY example you might see on a solving inequalities worksheet or problem set is the expression 2x + 3 [less than or equal to] 11. You can solve for x just like you would for an equality, or normal equation, by trying to get x alone. This is an example of two-step inequalities, or multi-step inequalities. The first step is to subtract the constant 3 from both sides, in order to move it to the right hand side, so that you have 2x [less than or equal to] 8. The second step is to get x alone on the left side by dividing both sides by 2 so that your answer for the inequality is x [less than or equal to] 4. In this case, solving the inequality used the exact same steps you would use to solve an "equal sign" equation.

Sometimes the steps are different from solving normal math equations. Solving for x is the same as solving a linear equality (-2x + 3 = 11) except that if you ever have to divide by a negative number or multiply by a negative number while solving for x, you must flip the sign, or switch the direction of your inequality symbol less than becomes greater than, for example). Here are the two cases of inequality problems WHEN YOU NEED TO REVERSE THE SIGN of the inequality because you're multiplying or dividing both sides of the inequality by a negative number:

2) DIVIDING EXAMPLE: For an inequality like -2x + 3 [less than or equal to] 11, after moving the constant 3 to the right side, you will need to divide both sides by negative 2 in order to get x alone on the left side for the solution. WHEN YOU DIVIDE BY A NEGATIVE NUMBER, you need to reverse the direction of the inequality symbol in your expression. In this example, after you divide out -2, you get x [greater than or equal to] -4 (instead of x less than or equal to -4) for the final solution. This example was also a two-step inequality problem.

3) MULTIPLYING EXAMPLE: For examples like (-1/2)x + 4 [greater than] 7, when you've moved the constant to the right side and have the inequality (-1/2)x [greater than] 3, the fastest way to get x alone is to clear the fraction by multiplying both sides by -2. This will cancel the 2 in the denominator and remove the negative sign in front of x on the left side. But WHEN YOU MULTIPLY BY A NEGATIVE, you have to flip the direction of the inequality symbol so that your answer is x [less than] -6 instead of x [greater than] -6. This problem was also a multi-step inequality.

Note on inequality rules: you don't have to switch the symbol if you're adding a negative number or subtracting a negative number: switching the inequality sign is only for when you're multiplying or dividing by a negative number.

MIT grad shows an easy way to use the Quotient Rule to differentiate rational functions and a shortcut to remember the formula. The calculus Quotient Rule derivative rule is one of the derivative rules for differentiation. It's used to take the derivative of a rational function. To skip ahead: 1) For an easy way to remember the Quotient Rule formula, skip to time 0:21. 2) For an example of how to use the Quotient Rule to take the derivative of a fraction or quotient of functions (rational function), skip to 1:41. This video is a basic introduction to the Quotient Rule for taking derivatives in calculus. Nancy formerly of MathBFF explains the steps.

For more help with Quotient Rule derivatives and HOW TO TAKE THE DERIVATIVE of a function using the DERIVATIVE RULES (Power Rule, Product Rule, Quotient Rule), jump to: https://youtu.be/QqF3i1pnyzU

The Quotient Rule (calculus) tells you how to find the derivative of rational functions (a fraction, or one function divided by another function). The formal definition (textbook definition) of the Quotient Rule is often unnecessarily complex and intimidating.

There is a memory trick, or mnemonic, for how to remember the Quotient Rule formula. All you need to remember is the song "LO dee-HI minus HI dee-LO, over LO LO," where "dee" means the "derivative of." "HI" means your top function in the numerator, and "LO" means your bottom function in the denominator.

In other words, multiply the bottom function times the derivative of the top function MINUS the top function times the derivative of the bottom function, then DIVIDED by the bottom function times itself. After you differentiate the function with the Quotient Rule, remember to simplify the expression as much as possible using algebra.

This video is a basic intro to the Quotient Rule.

For more help with Quotient Rule derivatives and HOW TO TAKE THE DERIVATIVE of a function using the DERIVATIVE RULES (Power Rule, Product Rule, Quotient Rule), jump to: https://youtu.be/QqF3i1pnyzU

The Quotient Rule (calculus) tells you how to find the derivative of rational functions (a fraction, or one function divided by another function). The formal definition (textbook definition) of the Quotient Rule is often unnecessarily complex and intimidating.

There is a memory trick, or mnemonic, for how to remember the Quotient Rule formula. All you need to remember is the song "LO dee-HI minus HI dee-LO, over LO LO," where "dee" means the "derivative of." "HI" means your top function in the numerator, and "LO" means your bottom function in the denominator.

In other words, multiply the bottom function times the derivative of the top function MINUS the top function times the derivative of the bottom function, then DIVIDED by the bottom function times itself. After you differentiate the function with the Quotient Rule, remember to simplify the expression as much as possible using algebra.

This video is a basic intro to the Quotient Rule.

MIT grad shows how to do synthetic division, a shortcut for long division. It's a fast way of dividing polynomials, if you're dividing one polynomial by a linear expression like x+1 or x-3. To skip ahead: 1) For how to know WHICH NUMBERS to put in the corner of the division box and the first row, skip to time 0:22. 2) For the synthetic division STEPS of carrying down coefficients, multiplying, and adding, skip to 1:12. 3) For how to WRITE the FINAL ANSWER and how to WRITE the REMAINDER (if the last number you get is not zero, there is a remainder), skip to 3:37. 4) For what to do if you have a MISSING TERM and putting a zero as its placeholder, skip to 5:04. Nancy formerly of MathBFF explains the steps.

For my video on how to do LONG DIVISION of polynomials instead, jump to: https://youtu.be/RPXMBIFG_W4

For my video on FACTORING to solve for the zeros of a polynomial, jump to: https://youtu.be/Z5MnP9da4EM

SYNTHETIC DIVISION is often faster than long division for dividing polynomials, if the polynomial you're dividing by is "x plus a number" or "x minus a number". It can be used to factor a polynomial and to find the zeros or roots of a polynomial.

HOW TO DO synthetic division of polynomials:

First draw a little corner symbol. Inside the corner, write the number that makes your denominator equal to zero. If your denominator is "x+1", you would write -1 in the corner. Then write the coefficients of your top polynomial in a row, to the right of the corner symbol. Leave space for another row and draw a horizontal line beneath that. Here are the steps to repeat for synthetic division, to carry down coefficient numbers, multiply, and add:

1) Drop down the first coefficient number.

2) Multiply this number by the corner constant.

3) Write the product in the second column, second row spot.

4) Add the two numbers in the second column, and write the sum below the line in the second column.

5) Repeat those steps until you have written a number in the final column spot below the line.

6) Write your new polynomial answer using this bottom row of numbers. The first number is the leading coefficient of your polynomial and is the coefficient of an x-term that is one degree less than your original polynomial. The second number is the coefficient of the second term, etc. For example, if the numbers are 1 4 3 2, the new polynomial is 1x^2 + 4x + 3 + 2/(x+1). Notice the last number is part of the remainder, so you write that number, 2, over the original divisor (x+1) as a fraction, for the remainder term.

Sometimes there is no remainder in synthetic division examples. If the last number you get is zero, in your final division numbers, then there is no remainder. Your final answer will just have a polynomial and no fraction term added at the end. Synthetic division is a way of factoring polynomials: if there is no remainder when you finish, you have found a true factor of the original polynomial. In the example above, if the remainder had been zero instead, the original polynomial would factor into (x+1)(x^2 + 4x + 3).

Note: also watch out for a missing term in your original polynomial: if there is no x-squared term, or no x term, etc, then in order to get the right answer for synthetic division, you will need to make sure to put a zero number in your first row of coefficient numbers, as a placeholder for that missing term.

For my video on how to do LONG DIVISION of polynomials instead, jump to: https://youtu.be/RPXMBIFG_W4

For my video on FACTORING to solve for the zeros of a polynomial, jump to: https://youtu.be/Z5MnP9da4EM

SYNTHETIC DIVISION is often faster than long division for dividing polynomials, if the polynomial you're dividing by is "x plus a number" or "x minus a number". It can be used to factor a polynomial and to find the zeros or roots of a polynomial.

HOW TO DO synthetic division of polynomials:

First draw a little corner symbol. Inside the corner, write the number that makes your denominator equal to zero. If your denominator is "x+1", you would write -1 in the corner. Then write the coefficients of your top polynomial in a row, to the right of the corner symbol. Leave space for another row and draw a horizontal line beneath that. Here are the steps to repeat for synthetic division, to carry down coefficient numbers, multiply, and add:

1) Drop down the first coefficient number.

2) Multiply this number by the corner constant.

3) Write the product in the second column, second row spot.

4) Add the two numbers in the second column, and write the sum below the line in the second column.

5) Repeat those steps until you have written a number in the final column spot below the line.

6) Write your new polynomial answer using this bottom row of numbers. The first number is the leading coefficient of your polynomial and is the coefficient of an x-term that is one degree less than your original polynomial. The second number is the coefficient of the second term, etc. For example, if the numbers are 1 4 3 2, the new polynomial is 1x^2 + 4x + 3 + 2/(x+1). Notice the last number is part of the remainder, so you write that number, 2, over the original divisor (x+1) as a fraction, for the remainder term.

Sometimes there is no remainder in synthetic division examples. If the last number you get is zero, in your final division numbers, then there is no remainder. Your final answer will just have a polynomial and no fraction term added at the end. Synthetic division is a way of factoring polynomials: if there is no remainder when you finish, you have found a true factor of the original polynomial. In the example above, if the remainder had been zero instead, the original polynomial would factor into (x+1)(x^2 + 4x + 3).

Note: also watch out for a missing term in your original polynomial: if there is no x-squared term, or no x term, etc, then in order to get the right answer for synthetic division, you will need to make sure to put a zero number in your first row of coefficient numbers, as a placeholder for that missing term.

Nancy shows how to solve any quadratic equation by factoring. To skip to the shortcut trick, go to time 6:11. Nancy formerly of mathbff explains the steps.

The shortcut trick ("The Magic X") helps you factor any tough quadratic that doesn't begin with x^2 but instead begins w/ 2x^2 or 3x^2, or 4x^2, etc, so you can then solve.

1) IF QUADRATIC STARTS WITH X^2: It's faster to use the normal method for factoring in this case: trial and error. Ex: x^2 + 4x - 12. Find 2 numbers that multiply to the last number, -12, AND that add to the second coefficient, positive 4. First make a list of all pairs of #s that multiply to -12. Then check which pair also adds to 4. Write factors & solve by setting each = 0. Solve for x.

2) IF QUADRATIC STARTS WITH 2X^2 OR 3X^2 OR 4X^2, ETC:

A) First check if leading coefficient (2 or 3 or 4, etc) is an overall constant you can factor out of every term. If it is, factor it out first, then use Method #1 above to factor the X^2 expression that's left. Set factors = 0 & solve.

B) If a constant can't be factored out evenly from every term, it'll be faster & easier to use shortcut "magic X" method instead of Method #1. See it explained at time 6:11 in video. Set factors = 0 & solve.

The shortcut trick ("The Magic X") helps you factor any tough quadratic that doesn't begin with x^2 but instead begins w/ 2x^2 or 3x^2, or 4x^2, etc, so you can then solve.

1) IF QUADRATIC STARTS WITH X^2: It's faster to use the normal method for factoring in this case: trial and error. Ex: x^2 + 4x - 12. Find 2 numbers that multiply to the last number, -12, AND that add to the second coefficient, positive 4. First make a list of all pairs of #s that multiply to -12. Then check which pair also adds to 4. Write factors & solve by setting each = 0. Solve for x.

2) IF QUADRATIC STARTS WITH 2X^2 OR 3X^2 OR 4X^2, ETC:

A) First check if leading coefficient (2 or 3 or 4, etc) is an overall constant you can factor out of every term. If it is, factor it out first, then use Method #1 above to factor the X^2 expression that's left. Set factors = 0 & solve.

B) If a constant can't be factored out evenly from every term, it'll be faster & easier to use shortcut "magic X" method instead of Method #1. See it explained at time 6:11 in video. Set factors = 0 & solve.

MIT grad shows how to factor quadratic expressions. If you want to skip to the shortcut method, jump to 5:06. Nancy formerly of mathbff explains the steps.

The shortcut method ("The Magic X") helps you factor any tough quadratic that doesn't begin with x^2 but instead begins with 2x^2 or 3x^2, or 4x^2, etc.

1) IF YOUR QUADRATIC STARTS WITH X^2: It is faster to use the normal method for factoring in this case: trial and error. Say you have x^2 + 4x - 12. You need to find two numbers that multiply to the last number, -12, AND that add to the second coefficient, positive 4. First make a list of all pairs of numbers that multiply to -12. Then check which pair also adds to positive 4. Use the two numbers in this pair to write your factors.

2) IF YOUR QUADRATIC STARTS WITH 2X^2 OR 3X^2 OR 4X^2, ETC:

A) First check whether your leading coefficient (2 or 3 or 4, etc) is just an overall constant that you can factor out of every term in the quadratic. If it is, factor it out first and then use Method #1 above to factor the X^2 expression that remains.

B) If your leading coefficient cannot be factored out evenly from every term, it will be faster and easier to use the shortcut "magic X" method instead of Method #1. See this method explained at time 5:06 in this video.

For how to SOLVE QUADRATIC EQUATIONS by factoring, jump to the video: https://youtu.be/Z5MnP9da4EM

The shortcut method ("The Magic X") helps you factor any tough quadratic that doesn't begin with x^2 but instead begins with 2x^2 or 3x^2, or 4x^2, etc.

1) IF YOUR QUADRATIC STARTS WITH X^2: It is faster to use the normal method for factoring in this case: trial and error. Say you have x^2 + 4x - 12. You need to find two numbers that multiply to the last number, -12, AND that add to the second coefficient, positive 4. First make a list of all pairs of numbers that multiply to -12. Then check which pair also adds to positive 4. Use the two numbers in this pair to write your factors.

2) IF YOUR QUADRATIC STARTS WITH 2X^2 OR 3X^2 OR 4X^2, ETC:

A) First check whether your leading coefficient (2 or 3 or 4, etc) is just an overall constant that you can factor out of every term in the quadratic. If it is, factor it out first and then use Method #1 above to factor the X^2 expression that remains.

B) If your leading coefficient cannot be factored out evenly from every term, it will be faster and easier to use the shortcut "magic X" method instead of Method #1. See this method explained at time 5:06 in this video.

For how to SOLVE QUADRATIC EQUATIONS by factoring, jump to the video: https://youtu.be/Z5MnP9da4EM

Nancy shows a surefire way to find the domain of any function. To skip ahead: 1) For POLYNOMIAL only, skip to time 0:45. 2) For FRACTION only, skip to time 1:44. 3) For SQUARE ROOT only, skip to 5:08. 4) For SQUARE ROOT ON BOTTOM of Fraction, skip to 8:44. 5) For SQUARE ROOT ON TOP of Fraction, skip to 10:09. Nancy formerly of mathbff explains the steps.

For Domain, you just need to watch out for Square Roots and/or Fractions!

If your function is...

1) a POLYNOMIAL (no square roots or fractions), i.e. f(x) = x^2 + 3x + 1, domain is "all real numbers."

2) a FRACTION (w/ no square root), i.e. f(x) = (2x+1)/(x^2+5x+6). Set bottom "not equal" to zero. Solve for x.

3) a SQUARE ROOT (w/ no fraction), i.e. f(x) = sqrt(x+1). Set part under root "greater than or equal to zero".

4) a FRACTION w/ A SQUARE ROOT ON BOTTOM, i.e. f(x) = (x^2+2x+3)/(sqrt(x+1)). Set part under root "greater than zero." Solve the inequality. NOTE THE DIFFERENCE: You must set it greater than zero, not "greater than OR EQUAL TO zero" because we can't have zero (or square root of zero) in the denominator.

5) a FRACTION w/ A SQUARE ROOT ON TOP, i.e. f(x) = (sqrt(x+1))/(x^2+4), i) Set the bottom "not equal to" zero & solve; ii) Set what's under the root "greater than or equal to zero" & solve; iii) Take intersection of your two solutions.

For Domain, you just need to watch out for Square Roots and/or Fractions!

If your function is...

1) a POLYNOMIAL (no square roots or fractions), i.e. f(x) = x^2 + 3x + 1, domain is "all real numbers."

2) a FRACTION (w/ no square root), i.e. f(x) = (2x+1)/(x^2+5x+6). Set bottom "not equal" to zero. Solve for x.

3) a SQUARE ROOT (w/ no fraction), i.e. f(x) = sqrt(x+1). Set part under root "greater than or equal to zero".

4) a FRACTION w/ A SQUARE ROOT ON BOTTOM, i.e. f(x) = (x^2+2x+3)/(sqrt(x+1)). Set part under root "greater than zero." Solve the inequality. NOTE THE DIFFERENCE: You must set it greater than zero, not "greater than OR EQUAL TO zero" because we can't have zero (or square root of zero) in the denominator.

5) a FRACTION w/ A SQUARE ROOT ON TOP, i.e. f(x) = (sqrt(x+1))/(x^2+4), i) Set the bottom "not equal to" zero & solve; ii) Set what's under the root "greater than or equal to zero" & solve; iii) Take intersection of your two solutions.

Nancy shows how to find the horizontal asymptote (of a rational function) with a quick and easy rule. Nancy formerly of mathbff explains the steps.

For how to find VERTICAL asymptotes instead, jump to the video: https://youtu.be/V137qmDN9Qw

The degree of a function is the highest power of x that appears in the polynomial. To find the horizontal asymptote, there are three easy cases. 1) If the degree of the numerator expression is less than the degree of the denominator expression, then the horizontal asymptote is y=0 (the x-axis). 2) If the degree of the numerator is equal to the degree of the denominator, then you can find the horizontal asymptote by dividing the first, highest term of the numerator by the first, highest term of the denominator. This will simplify to y = some constant (just a number). 3) If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote.

For how to find VERTICAL asymptotes instead, jump to the video: https://youtu.be/V137qmDN9Qw

The degree of a function is the highest power of x that appears in the polynomial. To find the horizontal asymptote, there are three easy cases. 1) If the degree of the numerator expression is less than the degree of the denominator expression, then the horizontal asymptote is y=0 (the x-axis). 2) If the degree of the numerator is equal to the degree of the denominator, then you can find the horizontal asymptote by dividing the first, highest term of the numerator by the first, highest term of the denominator. This will simplify to y = some constant (just a number). 3) If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote.

Nancy explains how to do long division with polynomials. Here I show clear steps to divide two polynomials using long division. I give two examples, one basic example as an introduction to the steps, and one more advanced example. Nancy formerly of mathbff explains the steps.

The steps for long division of polynomials are the same as for long division of numbers and can be summarized as:

1) Find how many times the first term of your divisor (on the left) goes into the first term of your dividend (on the right).

2) Write that number (or "x" expression) on the top line above your dividend, aligned vertically with the term it's most similar to (like term).

3) Multiply that number (or "x" expression) by each term in your divisor (on the left) and write that result on the line directly below your dividend.

4) Subtract the second line from the first line and write your result below.

5) Drop down another term from your dividend into your bottom line.

6) Find how many times the first term of your divisor goes into the first term of your bottom line and write that result on your top line above the dividend, aligned vertically with the term like it.

7) Repeat steps #3-6 until the first term of your divisor cannot go into the first term of your bottom line result anymore. This bottom line expression is your remainder.

8) Your answer is your top line result plus your remainder over the divisor.

The steps for long division of polynomials are the same as for long division of numbers and can be summarized as:

1) Find how many times the first term of your divisor (on the left) goes into the first term of your dividend (on the right).

2) Write that number (or "x" expression) on the top line above your dividend, aligned vertically with the term it's most similar to (like term).

3) Multiply that number (or "x" expression) by each term in your divisor (on the left) and write that result on the line directly below your dividend.

4) Subtract the second line from the first line and write your result below.

5) Drop down another term from your dividend into your bottom line.

6) Find how many times the first term of your divisor goes into the first term of your bottom line and write that result on your top line above the dividend, aligned vertically with the term like it.

7) Repeat steps #3-6 until the first term of your divisor cannot go into the first term of your bottom line result anymore. This bottom line expression is your remainder.

8) Your answer is your top line result plus your remainder over the divisor.

Nancy explains how to find the vertex of a parabola. This is a simple, fast way to identify the vertex, taken either from the equation of a parabola in "standard form," or from the equation in "vertex form." Nancy formerly of mathbff explains the steps.

The equation of a parabola is a quadratic equation and can come in either standard or vertex form. For each form, I give you an easy method to calculate the parabola vertex point. As a separate question, if you want to be able to convert your standard form into vertex form with a perfect square in your equation, you will need to "complete the square" first to get your parabola equation into vertex form.

Graphically, in geometry the vertex is either the lowest point on the parabola (the minimum) if the parabola is opening up, or it is the highest point on the parabola (the maximum) if the parabola is opening down.

The equation of a parabola is a quadratic equation and can come in either standard or vertex form. For each form, I give you an easy method to calculate the parabola vertex point. As a separate question, if you want to be able to convert your standard form into vertex form with a perfect square in your equation, you will need to "complete the square" first to get your parabola equation into vertex form.

Graphically, in geometry the vertex is either the lowest point on the parabola (the minimum) if the parabola is opening up, or it is the highest point on the parabola (the maximum) if the parabola is opening down.