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

##### 1 Answer

Jan 16, 2016

#### Explanation:

Find the point the normal line will intercept:

#f(0)=0#

The normal line will intercept the point

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

Differentiate

#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

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

The function and normal line graphed:

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