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

Jan 16, 2016

$y = - \frac{1}{4} x$

#### Explanation:

Find the point the normal line will intercept:

$f \left(0\right) = 0$

The normal line will intercept the point $\left(0 , 0\right)$.

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

Differentiate $f \left(x\right)$:

$f ' \left(x\right) = 4 - 2 x$

The slope of the tangent line is:

$f ' \left(0\right) = 4$

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

Thus, the slope of the normal line is $- \frac{1}{4}$.

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

$y = - \frac{1}{4} x$

The function and normal line graphed:

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