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

Apr 25, 2018

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

#### Explanation:

$f \left(x\right) = {\left(2 x - 2\right)}^{2} + 4 x$
$f ' \left(x\right) = 2 \left(2 x - 2\right) \times 2 + 4$
$f ' \left(x\right) = 4 \left(2 x - 2\right) + 4$

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

So the gradient of the normal is $\frac{1}{28}$ as the two gradients must equal to $- 1$ for perpendicular lines

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

$\left(y - 52\right) = \frac{1}{28} \left(x + 3\right)$

$28 y - 1456 = x + 3$

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

Apr 25, 2018

$y = \left(\frac{1}{28}\right) x + 52.11$

#### 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# ; $f \left(x\right) = 52$ at $x = - 3$
$y = \left(\frac{1}{28}\right) x + b$ ; $52 = \left(\frac{1}{28}\right) \left(- 3\right) + b$ ; $b = 52.11$

The final line equation is thus:
$y = \left(\frac{1}{28}\right) x + 52.11$