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

##### 1 Answer
Nov 13, 2015

$y = x - 2$

#### Explanation:

Firstly, $f \left(2\right) = 2 - {2}^{2} / 2 = 0$.

Thus the point $\left(2 , 0\right)$ lies on the graph of f at the point 2.

Now the derivative $f ' \left(x\right) = 1 - x$,
and evaluated at the point 2 we get $f ' \left(2\right) = 1 - 2 = - 1$.

This means that the gradient of the function f at the point when x=-2 is -1.

Now a line normal at this point is perpendicular and will hence have a gradient of 1 since the product of gradients of perpendicular lines must be -1.

Bu t a normal line is still a straight line so has a linear equation of form $y = m x + c$.
If we substitute the point (2,0) and the gradient 1 into this linear equation we get
$0 = 2 + c \implies c = - 2$.

Therefore the equation of the required normal line is $y = x - 2$.