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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 1: AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c
- 2011 Calculus AB free response #1a
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

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# 2015 AP Calculus AB 6a

Equation for tangent line.

## Want to join the conversation?

- Can you use point-slope form for the equation at0:35?(3 votes)
- Yes, and on the AP Exam you wouldn't even need to simplify the equation. Your final answer could be

y-1 = 1/4(x+1) and that would be acceptable.(1 vote)

- what confuses me a lot is that sal says "this line is tangent to the curve. So includes this point and only that point. "at1:34but think tangent line is just secant line when the tow points are veryyyyyyyyy near to each other.(1 vote)

## Video transcript

- [Voiceover] Consider the
curve given by the equation Y to the third minus XY is equal to two. It can be shown that the derivative of Y with respect to X is equal
to Y over three Y squared minus X. All right? Write an equation for the
line tangent to the curve at the point negative one comma one. All right, so we can figure
out the equation for the line if we know the slope of the line and we know a point that it goes through so that should be enough to figure out the equation of the line. So the line's going to have a form Y is equal to MX plus B. M is the slope and is
going to be equal to DY/DX at that point, and we know that that's
going to be equal to. We'll see Y is, when X is negative one, Y is one, that sits on this curve. So Y is one. So one over three Y squared. So three times one squared which is three, minus X, when Y is one, X is negative one, or when
X is negative one, Y is one. So X is negative one here. And so this is the same thing as three plus positive one, and so this is equal to one fourth and so the equation of our line is going to be Y is equal
to one fourth X plus B. Now we need to solve for B and we know that point negative one comma one is on the line, so we can use that
information to solve for B. This line is tangent to the curve. So includes this point
and only that point. That's what it has in
common with the curve and so why is equal to one when X is equal to negative one, plus B and so we have one is equal to negative one fourth plus B. You add one fourth to both sides, you get B is equal to, we could either write it
as one and one fourth, which is equal to five fourths, which is equal to 1.25. We could write it any of those ways, so the equation for the
line tangent to the curve at this point is Y is equal to our slope is one fourth X plus and I could write it in any of these ways. I'll write it as plus five over four and we're done at least with
that part of the problem.