# What is the equation of the normal line of f(x)=5x^3+3x^2+x-5 at x=2?

Dec 31, 2015

Equation of normal is $y - 49 = - \frac{1}{73} \left(x - 2\right)$ This is in Slope-Point form.

#### Explanation:

To find the equation of normal line we need to find the slope of the normal line and a point through which the line passes through.

We are given

$f \left(x\right) = 5 {x}^{3} + 3 {x}^{2} + x - 5$ and we are asked to find equation at $x = 2$
First let us find $y$ when $x = 2$

$y = f \left(2\right) = 5 {\left(2\right)}^{3} + 3 {\left(2\right)}^{2} + \left(2\right) - 5$
$y = f \left(2\right) = 5 \left(8\right) + 3 \left(4\right) + 2 - 5$
$y = f \left(2\right) = 40 + 12 - 3$
$y = 49$

Now we have a point $\left(2 , 49\right)$

The slope of the tangent can be found by finding the derivative of $f \left(x\right)$ and evaluated at $x = 2$

$f \left(x\right) = 5 {x}^{3} + 3 {x}^{2} + x - 5$
$f ' \left(x\right) = 15 {x}^{2} + 6 x + 1$
$m = f ' \left(2\right) = 15 {\left(2\right)}^{2} + 6 \left(2\right) + 1$
$m = 15 \left(4\right) + 12 + 1$
$m = 60 + 12 + 1$
$m = 73$

Slope of normal ${m}_{\text{normal}} = - \frac{1}{m}$
Slope of normal ${m}_{\text{normal}} = - \frac{1}{73}$
Equation of a line passing through a point $\left({x}_{1} , {y}_{1}\right)$ and slope $m$ is given by

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Here we have the point $\left(2 , 49\right)$ and slope $- \frac{1}{73}$

Equation of normal is $y - 49 = - \frac{1}{73} \left(x - 2\right)$