Tangents of polynomials | Derivative rules | AP Calculus AB | Khan Academy

What you see here in blue, this is the graph of ( y ) is equal to ( f(x) ) where ( f(x) ) is equal to ( x^3 - 6x^2 + x - 5 ). What I want to do in this video is think about what is the equation of the tangent line when ( x ) is equal to 1, so we can visualize that.

So this is ( x ) equaling 1 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.

What we want to do is find the equation, the equation of that line. And if you are inspired, I encourage you to pause the video and try to work it out.

Well, the way that we could do this is if we find the derivative at ( x ) equals 1. 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. 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 evaluate ( f(1) ). So ( f(1) ) is equal to ( 1^3 ), which is 1, minus ( 6 \times 1^2 ), so it's just minus 6, and then plus 1 plus 1 minus 5. So this is equal to what? 2 minus 11, which is equal to negative 9.

And that looks about right; that looks like about negative 9 right over there. The scales are different on the y and the x-axis, and so that is ( f(1) ). It is negative 9. Did I do that right? This is negative 5. Negative? Yep, negative 9.

And now let's evaluate what the derivative is at 1. So what is ( f'(x) )? ( f'(x) )—well here it's just a polynomial—take the derivative of ( x^3 ). Well, we apply the power rule; we bring the 3 out front, so you get ( 3x^{2} ), and then we go one less than 3 to get the second power.

Then you have minus ( 6x^2 ), so you bring the 2 times 6 to get 12. So minus 12 ( x ) to the—well, 2 minus 1 is 1 power, so that's the same thing as ( 12x ), and then plus the derivative of ( x ) is just 1; that's just going to be 1.

And if you view this as ( x^{1} ), we're just bringing the 1 out front and decrementing the 1. So it's ( 1 \times x^{0} ), which is just 1. And then the derivative of a constant here is just going to be 0.

So this is our derivative of ( f ), and if we want to evaluate it at 1, ( f'(1) ) is going to be ( 3 \times 1^2 ), which is just 3, minus 12 times 1, so it's just minus 12, and then we have plus 1. So this is ( 3 - 12 ) is negative 9, plus 1 is equal to negative 8.

So we know the slope right over here is a slope of negative 8. We know a point on that line—it contains the point (1, -9), so we could use that information to find the equation of the line.

The line, just to remind ourselves, has the form ( y = mx + b ) where ( m ) is the slope. So we know that ( y ) is going to be equal to ( -8x + b ). 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 9—let me just write this here: ( y ) is equal to negative 9 when ( x ) is equal to 1. And so we get— we get negative 9 is equal to ( -8 \times 1 ), so negative 8 plus ( b ). Well, let's see, we could add 8 to both sides and we get negative 1 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 is ( y = -8x - 1 ).