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

Lesson 2: AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

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# 2015 AP Calculus BC 5a

Equation of tangent line.

## Want to join the conversation?

- Could I write the equation in point slope form?(4 votes)
- I think that's easier too(3 votes)

- If anyone still looks at these, is it alright to have an unsimplified answer or are points taken off for that?(1 vote)
- That depends on the person checking the answers. Some want a simplified answer, some don't mind. If I were you, I would simplify answers to stay on the safer side.(2 votes)

## Video transcript

- [Voiceover] Consider the
function f of x is equal to one over x-squared minus k-x,
where k is a nonzero constant. The derivative of f is given by, and they give us this
expression right over here. That's nice that they took
the derivative for us. Now part a, let k equals
three so that f of x is equal to one over x-squared minus three-x. So they set k equal to three. Write an equation for the line tangent to the graph of f at the point whose x-coordinate is four. So to find an equation for a line, the equation of a line
is gonna be in the form y is equal to m-x plus b, where m is the slope of the line and b is the y-intercept. And the slope of the
line, right over here, this needs to be equal to
the derivative evaluated when x is equal to four. So we could say, y is equal to... Let me write it this way. We could say that m is
going to be equal to f-prime when x is equal to four. So, f-prime of four, which is equal to... Well, we know that k is equal to three. So, they gave us f-prime of x. So, it's going to be
three minus two times, we're taking f-prime of four, so minus two times four over x-squared. So that's going to be
four-squared, minus k, is three times four, three times four, and then
we square that whole thing. And so what is this going to be? This is an eight right over here, and all I did is f-prime of
x, when k is equal to three, is gonna be three minus
two-x over x-squared minus three-x and all of that squared. And I want to evaluate
what f-prime of four is. So, every place where I saw an x, I substituted with a four. Where I saw the k, k is three. And so this is going to be equal to... The numerator is three minus
eight, is negative five, over this is 16 minus 12, which is going to be four, 16 minus 12 is four,
and then we square it. So it's gonna be negative five-sixteenths. And so let me write it this way, m is equal to negative five-sixteenths. So, how do we figure out b now? Well, what are the coordinates? When x is equal to four, what
is y going to be equal to? Well, y is equal to f of x. So we know that y, on
the curve, we know that y is going to be equal to f of four. So before we evaluated f-prime of four. Now we're going to evaluate
y as being f of four which is equal to one over four-squared, four-squared minus three times four, and so that is equal to one over 16 minus 12, which is four. And so this point, right
here, when x is four, then y is equal to one-fourth. So, we can use that
information to solve for b. When y is one-fourth, x... So, we're gonna say if y is equal to m, which is negative five-sixteenths, negative five-sixteenths, times x. Well, when y is one-fourth, x is... Or when I say when x is equal to four, y is one-fourth, and then plus b. So I can now solve for b. So all I did is I used f-prime of x. I used f-prime of x to figure out m, when x is equal to four. And then, I said, "Okay,
well what is the value of y "when x is equal to four?" And so, if I know y, m, and
x, then I can solve for b. And so, let's just do that. We get one-fourth is equal to four times negative five-sixteenths,
is negative five over four, plus b. I can add five-fourths to both sides, and I get five-fourths plus
one-fourth is equal to b, or b is equal to six-fourths, six over four, which you can say... Well, there's a bunch of
ways you could write this. We could just say this
is just equal to 1.5. And so our equation is y
is equal to negative five over 16 x plus 1.5. Or if we wanted to write
everything as a fraction, we could say y is equal to negative five over 16 x plus three-halves. Six-fourths is the same thing as three-halves, and there you go.