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## Polynomial functions differentiation

# Tangents of polynomials

AP.CALC:

FUN‑3 (EU)

, FUN‑3.A (LO)

, FUN‑3.A.3 (EK)

## Video transcript

- [Voiceover] What you see here in blue this is the graph of Y is equal to F of X. Where F of X is equal to X to the third minus six X squared plus X minus five. What I want to do in this video is think about what is the equation of the tangent line when X is equal to one? So we can visualize that. So, this is X equaling
one right over here. This is the value of the function. When X is equal to one. Right over there. And then the tangent line looks something like will look something like. I know I can do a better job than that. It's going to look something like that. And what we want to do
is find the equation the equation of that line. And if you are inspired
I encourage you to be, pause the video and try to work it out. Well the way that we can do this is if we find the derivative at X equals one the derivative is the
slope of the tangent line. And so we'll know the
slope of the tangent line. And we know that it contains that point and then we can use that
to find the equation of the tangent line. So let's actually just, let's just. So we want the equation
of the tangent line when X is equal to one. So let's just first of
all evaluate F of one. So F of one is equal to one to the third power which is one. Minus six times ones squared, so it's just minus six. And then plus one plus one minus five. So, this is equal to what? Two minus 11? Which is equal to negative nine. And that looks about right. That looks like about negative
nine right over there. The scales are different
on the Y and the X axis. So that is F of one. It is negative nine. Did I do that right? This is negative five. Yep, negative nine. And now let's evaluate
what the derivative is at one. So, what is F prime of X? F prime of X. Well here it's just a polynomial. You take the derivative of X to the third while we apply the power rule. We bring the three out front. So you get three X to the. And then we go one less than three to get the second power. Then you have minus six X squared. So you bring the two times the six to get 12. So minus 12 X to the well two minus one is one power so that's the same thing as 12 X. And then plus the derivative of X is just one. That's just going to be one. And if you view this
as X to the first power we're just bringing the one out front and decrementing the one. So it's one times X to the zero power which is just one. And then the derivative of a constant here is just going to be zero. So this is our derivative of F and if we want to evaluate it at one F prime of one is going to be three times one squared which is just three minus 12 times one which is just minus 12. And then we have plus one. So this is three minus 12 is negative nine plus one is equal to negative eight. So we know the slope right over here is the slope of negative eight. We know a point on that line it contains the point one negative nine so we can use that information to find the equation of the line. The line, just to remind ourselves, has the four. Y is equal to M X plus B. Where M is the slope. So we know that Y is going to be equal to negative eight X plus B. And now we can substitute the X and Y value that we know sits on that line to solve for B. So we know that Y is
equal to negative nine. Let me just write this here. Y is equal to negative nine when X is equal to when X is equal to one. And so we get we get negative nine is equal to negative eight times one. So negative eight plus B. Well, let's see. We could add we could eight to both sides and we get negative one is equal to B. So we're done. The equation of the line the equation of this line that we have in magenta right over there that is that is Y is equal to the slope is negative eight X. And then the Y-intercept minus one.