Question

What is the equation of the line normal to `f(x)=(4-2x)^2` at `x=-3`?

Answer 1

`y-100=1/40(x+3)`

Explanation:

First, find the point the normal line will intercept.

`f(-3)=(4+6)^2=100`

The normal line will pass through the point `(-3,100)`.

To find the slope of the normal line, we must first know the slope of the tangent line. The slope of the tangent line can be found through calculating `f'(-3)`.

First, find `f'(x)`. Through the chain rule,

`f'(x)=2(4-2x)*(-2)=-4(4-2x)`

The slope of the tangent line is

`f'(-3)=-4(4+6)=-40`

However, the normal line is perpendicular to the tangent line. Perpendicular lines have opposite reciprocal slopes. Thus, the slope of the normal line is the opposite reciprocal of `-40` which is `1/40`.

Relate the point the normal line intercepts, `(-3,100)`, and the slope of the line, `1/40`, in an equation in point-slope form:

`y-100=1/40(x+3)`