Question

What is the equation of the normal line of `f(x)=5x^3+x^2-5x-2` at `x=2`?

Answer 1

`f(x) = 5x^3+x^2-5x-2`

At `x=2`, we have `y=f(2) = 44-12 = 32`

`f'(x) = 15x^2+2x-5`, so the slope of the tangent line is `f'(2) = 64-5 = 59`

The normal line is perpendicular to the tangent, so its slope is `-1/59`

An equation for the normal line is `y=32-1/59(x-2)`.

This can also be written `y=-1/59x+32 2/59`.