# What is the equation of the normal line of f(x)=2x^3-8x^2+2x-1 at x=-1?

Feb 9, 2016

$y + 13 = - \frac{1}{24} \left(x + 1\right)$

#### Explanation:

First, find the point the normal line will intercept by finding the function value at $x = - 1$.

f(-1)=2(-1)^3-8(-1)^2+2(-1)-1=-2-8-2-1=ul(-13

The normal line will pass through the point $\left(- 1 , - 13\right)$.

Before we can find the slope of the normal line, we must first find the slope of the tangent line. The slope of the tangent line is equal to the value of the function's derivative at $x = - 1$.

Find the derivative of the function through the power rule:

$f \left(x\right) = 2 {x}^{3} - 8 {x}^{2} + 2 x - 1$

$f ' \left(x\right) = 6 {x}^{2} - 16 x + 2$

The slope of the tangent line is

f'(-1)=6(-1)^2-16(-1)+2=6+16+2=ul(24

Since the tangent line and normal line are perpendicular, their slopes will be opposite reciprocals. The opposite reciprocal of $24$ is $- \frac{1}{24}$.

We can relate the information we know about the normal line as a linear equation in point-slope form, which takes a point $\left({x}_{1} , {y}_{1}\right)$ and slope $m$:

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

Since the normal line passes through $\left(- 1 , - 13\right)$ and has slope $- \frac{1}{24}$, its equation is

$y + 13 = - \frac{1}{24} \left(x + 1\right)$

Graphed are the function and its normal line:

graph{(2x^3-8x^2+2x-1-y)(y+13+(x+1)/24)=0 [-5, 7, -18.16, 2.12]}