# How do you find the equation of the tangent and normal line to the curve y=x^2 at x=-3?

Aug 26, 2016

Equation of the tangent

$y = - 6 x - 9$

Equation of the Normal

$y = \frac{1}{6} x + \frac{19}{2}$

#### Explanation:

Given -

$y = {x}^{2}$

It is a quadratic function. The curve is a upward facing parabola.

Its first derivative gives the slope at any given point on the curve.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

Slope exactly at $x = - 3$

${m}_{1} = 2 \left(- 3\right) = - 6$

At x=-3 ; y=(-3)^2=9

The tangent and Normal are passing through the point $\left(- 3 , 9\right)$

Slope of the tangent is ${m}_{2} = - 6$

Steps to get Equation of the tangent -

$m x + c = y$

$\left(- 6\right) \left(- 3\right) + c = 9$

$18 + c = 9$

$c = 9 - 18 = - 9$

Equation of the tangent is -

$y = - 6 x - 9$

Steps to get Equation of the Normal -

Normal's slope ${m}_{3} = - \left(\frac{1}{{m}_{2}}\right) = - \left(\frac{1}{- 6}\right) = \frac{1}{6}$

$m x + c = y$
$\frac{1}{6} \left(- 3\right) + c = 9$
$- \frac{1}{2} + c = 9$
$c = 9 + \frac{1}{2} = \frac{18 + 1}{2} = \frac{19}{2}$

Equation of the Normal -

$y = \frac{1}{6} x + \frac{19}{2}$