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

Feb 2, 2016

$y = - \frac{1}{5} x - \frac{9}{5}$

#### Explanation:

To build the equation of the normal line, you need a point on the normal line and its slope.

1) Point on the normal line

As you are searching for the normal line at $x = 1$, your $x$ coordinate is $x = 1$.

To find the $y$ value of the point, you need to evaluate $f \left(x\right)$ for $x = 1$:

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

Thus, the point is $P \left(1 | - 2\right)$

2) Slope of the normal line

To compute the slope of the normal line, first of all, you need the derivative of $f \left(x\right)$:

$f ' \left(x\right) = - 4 {x}^{3} + 6 {x}^{2} + 4 x - 1$

Now, you need to evaluate the derivative at $x = 1$. This will give you the slope of the tangent at $x = 1$.

$f ' \left(1\right) = - 4 \cdot {1}^{3} + 6 \cdot {1}^{2} + 4 \cdot 1 - 1 = - 4 + 6 + 4 - 1 = 5$

Thus, the slope of the tangent at $x = 1$ is ${m}_{t} = 5$.

We know that the slope of the normal line ${m}_{n}$ can be computed as

${m}_{n} = - \frac{1}{m} _ t = - \frac{1}{5}$

3) Computing the equation of the normal line

Now that we have the point and the slope, we can build the equation.

Any line has the form

$y = m \cdot x + n$

We already have $y = - 2$, $m = - \frac{1}{5}$ and $x = 1$. Let's find $n$:

$- 2 = - \frac{1}{5} \cdot 1 + n$

$- 2 = - \frac{1}{5} + n$

... add $\frac{1}{5}$ on both sides...

$- \frac{9}{5} = n$

Thus, the equation of the normal line is

$y = - \frac{1}{5} x - \frac{9}{5}$