Question

How do you find the tangent line to the curve y=x³−9x at the point where x=1?

Answer 1

Given: y=x3−9x.

Let f(x)=y=x³−9x. At x=1, f(x)=13−9⋅1=1−9=−8.

So, the point we are targeting is (1,−8).

To find the slope of the tangent line there, we must differentiate f(x) and then plug in x=1 there.

∴f'(x)=3x²−9

At x=1,

f'(1)=3⋅12−9

      =3−9

     =−6

So, the slope of the tangent line is −6.

Now, we use the point-slope formula to compute the equation, that is,

y−y0=m(x−x0)

(x0,y0) are the original coordinates

Therefore, we get,

y−(−8)=−6(x−1)

y+8=−6(x−1)

y+8=−6x+6

y=−6x+6−8

=−6x−2

:.Tangent line y+6x+2=0.