# What is the equation of the normal line of f(x)=2x^3-x^2-3x+9 at x=-1?

Jun 19, 2016

Let's first find what point the tangent passes through, given $x = a$.

#### Explanation:

$f \left(x\right) = 2 {\left(- 1\right)}^{3} - {\left(- 1\right)}^{2} - 3 \left(- 1\right) + 9$

$f \left(x\right) = - 2 - 1 + 3 + 9$

$f \left(x\right) = 9$

Therefore, the tangent passes through $\left(- 1 , 9\right)$.

Now that we know this, we must differentiate the function.

By the power rule:

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

The slope of the tangent is given by evaluating $f \left(a\right)$ inside the derivative, a being $x = a$.

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

$f ' \left(- 1\right) = 6 + 2 - 3$

$f ' \left(- 1\right) = 5$

The slope of the tangent is $5$. The normal is always perpendicular to the tangent, so the slope will be the negative reciprocal of that of the tangent.

This means the slope of the normal line is $- \frac{1}{5}$. By point slope form, we can find the equation of the normal line:

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

$y - 9 = - \frac{1}{5} \left(x - \left(- 1\right)\right)$

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

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

$y = - \frac{1}{5} x + \frac{44}{5}$

$\therefore$ The equation of the normal line is $y = - \frac{1}{5} x + \frac{44}{5}$.

Hopefully this helps!