Question

What is the equation of the normal line of `f(x)=x^2` at `x=-5`?

Answer 1

`y=-10x-25`

Explanation:

The equation of a straight line is determined by computing its `color(red)(slope)` and the point it passes through name it `(color(purple)(x_1,y_1))`

This line is normal to `f(x)` at `color(purple)(x=-5)` so it passes through the point`(color(purple)(-5,f(-5)))`

`color(purple)(f(-5))=(-5)^2=25`

This line passes through the point `(color(purple)(-5,25))`

The `color(red)(slope)` at `x=-5` is determined by computing `color(red)(f'(-5))`
`f(x)` is differentiated by using power rule differentiation
`(x^n)'=n(x^(n-1))`

`f'(x)=2x`
`color(red)(f'(-5)=2(-5)=-10)`

The equation is:

`y-y_1=color(red)(slope)(x-x_1)`

`y-f(-5)=-10(x-(-5))`

`y-25=-10(x+5)`

`y=-10x-50+25`

`y=-10x-25`