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

Nov 30, 2015

$y = - \frac{1}{8} x - \frac{33}{8}$

#### Explanation:

To compute the equation of the normal line, you need:

• the slope of the normal line
• one point that is on the normal line

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You already know that the normal line intercepts your function $f \left(x\right)$ at $x = - 1$.

So, the only thing left to do is computing the $y$ value for $x = - 1$:

$f \left(- 1\right) = - 2 \cdot \left(- 1\right) - 8 \cdot 1 - 2 \cdot \left(- 1\right) = 2 - 8 + 2 = - 4$

So, your point is $\left(- 1 | - 4\right)$.

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2) Now, let's take care of the slope . To do so, you need to differentiate $f \left(x\right)$ and then compute the value of $f ' \left(- 1\right)$.

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

$f ' \left(- 1\right) = - 6 \cdot 1 - 16 \cdot \left(- 1\right) - 2 = - 6 + 16 - 2 = 8$

This means that the slope of the tangent line at $x = - 1$ is ${m}_{t} = 8$.
The slope of the normal line is ${m}_{n} = - \frac{1}{m} _ t = - \frac{1}{8}$.

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3) We have the point and the slope. Let's compute the equation of the normal line.

The equation of any line is $y = m x + n$.

We know that $m = - \frac{1}{8}$ and for $x$ and $y$ we can plug the $x$ and $y$ values of the point that we know is on the line: $x = - 1$ and $y = - 4$.

With this knowledge, we can compute $n$:

$- 4 = - \frac{1}{8} \cdot \left(- 1\right) + n$

$\iff - 4 = \frac{1}{8} + n$

$\iff n = - \frac{33}{8}$

So, the equation of the normal line is

$y = - \frac{1}{8} x - \frac{33}{8}$

Hope that this helped!