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

Aug 3, 2016

$y = - \frac{1}{13} x + \frac{53}{13}$

#### Explanation:

Given -

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

The first derivative gives the slope at any given point

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

At $x = 1$ the slope of the curve is -

${m}_{1} = 8 \left({1}^{3}\right) + 12 \left({1}^{2}\right) - 4 \left(1\right) - 3$
${m}_{1} = 8 + 12 - 4 - 3 = 13$

This is the slope of the tangent drawn to the point $x = 1$ on the curve.

The y-coordinate at $x = 1$is

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

$y = 2 + 4 - 2 - 3 + 3 = 4$

The normal and the tangent are passing through the point $\left(1 , 4\right)$

The normal cuts this tangent vertically. Hence, its slope must be

${m}_{2} = - \frac{1}{13}$

[You must know the product of the slopes of the two vertical lines is ${m}_{1} \times {m}_{2} = - 1$ in our case $13 \times - \frac{1}{13} = - 1$

The equation of the normal is -

$- \frac{1}{13} \left(1\right) + c = 4$

$c = 4 + \frac{1}{13} = \frac{52 + 1}{13} = \frac{53}{13}$

$y = - \frac{1}{13} x + \frac{53}{13}$

Aug 3, 2016

$x + 13 y = 53$ or $y = - \frac{x}{13} + \frac{53}{13}$

#### Explanation:

$f \left(x\right) = 2 {x}^{4} + 4 {x}^{3} - 2 {x}^{2} - 3 x + 3$

To find the equation to the normal First step is to find the slope.

The first derivative of a curve at a particular point is the slope of the
tangent at that point.

Use this idea let us first find the slope of the tangent

$f ' \left(x\right) = 8 {x}^{3} + 12 {x}^{2} - 4 x - 3$

$f ' \left(1\right) = 8 + 12 - 4 - 3 = 13$

The slope of the tangent to the given curve at x=1 is 13

The product of the slopes of the tangent and normal would be -1 .
so the slope of the normal is $- \frac{1}{13.}$

we need to find f(x) at $x = 1 , f \left(1\right) = 2 + 4 - 2 - 3 + 3 = 4$
we have slope is $- \frac{1}{13}$ and the point is (1,1).

We have $m = - \frac{1}{13}$ and $\left(x 1 , y 1\right) \rightarrow \left(1 , 4\right)$

$y - 4 = \left(- \frac{1}{13}\right) \left(x - 1\right)$
$13 \left(y - 4\right) = \left(- 1\right) \left(x - 1\right)$
$13 y - 52 = - x + 53$
$x + 13 y = 53$