Question

What is the equation of the line normal to `f(x)=(2x-2)^2+4x` at `x=-3`?

Answer 1

`x-28y+1459=0` is the equation of the normal

Explanation:

`f(x)= (2x-2)^2+4x`
`f'(x)=2(2x-2)times2+4`
`f'(x)=4(2x-2)+4`

At `x=-3`, `f'(-3)= 4(-6-2)+4 = -28`
This is the gradient of the tangent.

So the gradient of the normal is `1/28` as the two gradients must equal to `-1` for perpendicular lines

Equation of normal 0f #(-3,52) :

`(y-52)=1/28(x+3)`

`28y-1456=x+3`

`x-28y+1459=0` is the equation of the normal

Answer 2

`y = (1/28)x + 52.11`

Explanation:

“Normal” means perpendicular to the line, or that the slopes will be opposite.
First we need to find the tangent line, which is the first derivative of the expression at x = -3.
https://www.symbolab.com/solver/tangent-line-calculator

Step-by-step solution here (long):
https://www.symbolab.com/solver/tangent-line-calculator/tangent%20of%20f%5Cleft(x%5Cright)%3D%5Cleft(2x%E2%88%922%5Cright)%5E%7B2%7D%2B4x%2C%20at%20x%3D%E2%88%923
`f(x) = -28x – 32`

THEN we need to find the equation of a line perpendicular to this one (inverse slope) passing through `x = -3`

`f(x) = -28x – 32` ; `f(x) = 52` at `x = -3`
`y = (1/28)x + b` ; `52 = (1/28)(-3) + b` ; `b = 52.11`

The final line equation is thus:
`y = (1/28)x + 52.11`