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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 11

Lesson 2: AP Calculus BC 2015

# 2015 AP Calculus BC 5a

Equation of tangent line.

## 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.