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

2 Answers
Apr 25, 2018

#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

Apr 25, 2018

#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#