Question

What is the equation of the normal line of `f(x)=4x-x^2` at `x=0`?

Answer 1

`y=-1/4x`

Explanation:

Find the point the normal line will intercept:

`f(0)=0`

The normal line will intercept the point `(0,0)`.

Before you can find the slope of the normal line, find the slope of the tangent line.

Differentiate `f(x)`:

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

The slope of the tangent line is:

`f'(0)=4`

Since the tangent line and normal line are perpendicular, they will have opposite reciprocal slopes.

Thus, the slope of the normal line is `-1/4`.

Since we know the line will pass through the origin, the equation of the normal line is:

`y=-1/4x`

The function and normal line graphed:

graph{(4x-x^2-y)(y+x/4)=0 [-10, 10, -5, 5]}