Question

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

Answer 1

`y=x-2`

Explanation:

Firstly, `f(2)=2-2^2/2=0`.

Thus the point `(2,0)` lies on the graph of f at the point 2.

Now the derivative `f'(x)=1-x`,
and evaluated at the point 2 we get `f'(2)=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=mx+c`.
If we substitute the point (2,0) and the gradient 1 into this linear equation we get
`0=2+c =>c=-2`.

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