What is the equation of the normal line of f(x)=2x^4+4x^3-2x^2-3x+3 at x=-1?

Aug 27, 2016

$y = - \frac{1}{11} x + \frac{21}{11}$

Explanation:

Given -

$y = 2 {x}^{4} + 4 {x}^{3} - 2 {x}^{2} - 3 x + 3$

The first derivative gives the slope of the curve at any given point.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 {x}^{3} + 12 {x}^{2} _ 4 x - 3$

Slope of the curve exactly at $x = - 1$

${m}_{1} = 8 {\left(- 1\right)}^{3} + 12 {\left(- 1\right)}^{2} - 4 \left(- 1\right) + 3$

${m}_{1} = - 8 + 12 + 4 + 3 = 11$

This is the slope of the tangent also.

Normal cuts the tangent vertically.

So its slope is

${m}_{2} = - \frac{1}{11}$ [Since; ${m}_{1} \times {m}_{2} = - 1$]

The normal passes through $x = - 1$

Find the Y-coordinate

$y = 2 {\left(- 1\right)}^{4} + 4 {\left(- 1\right)}^{3} - 2 {\left(- 1\right)}^{2} - 3 \left(- 1\right) + 3$
$y = 2 - 4 - 2 + 3 + 3 = 2$

The Normal passes through the point $\left(- 1 , 2\right)$

Then find the equation of the Normal

$m x + c = y$

$\left(- \frac{1}{11}\right) \left(- 1\right) + c = 2$
$\frac{1}{11} + c = 2$
$c = 2 - \frac{1}{11} = \frac{21}{11}$

The equation of the Normal is-

$y = - \frac{1}{11} x + \frac{21}{11}$