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

May 7, 2018

The normal line is given by $y = - x - 4$

#### Explanation:

Rewrite $f \left(x\right) = \frac{2 {x}^{2} + 1}{x}$ to $2 x + \frac{1}{x}$ to make differentiation simpler.

Then, using the power rule, $f ' \left(x\right) = 2 - \frac{1}{x} ^ 2$.

When $x = - 1$, the y-value is $f \left(- 1\right) = 2 \left(- 1\right) + \frac{1}{-} 1 = - 3$. Thus, we know that the normal line passes through $\left(- 1 , - 3\right)$, which we will use later.

Also, when $x = - 1$, instantaneous slope is $f ' \left(- 1\right) = 2 - \frac{1}{- 1} ^ 2 = 1$. This is also the slope of the tangent line.

If we have the slope to the tangent $m$, we can find the slope to the normal via $- \frac{1}{m}$. Substitute $m = 1$ to get $- 1$.

Therefore, we know that the normal line is of the form
$y = - x + b$

We know that the normal line passes through $\left(- 1 , - 3\right)$. Substitute this in:
$- 3 = - \left(- 1\right) + b$
$\therefore b = - 4$

Substitute $b$ back in to get our final answer:
$y = - x - 4$

You can verify this on a graph:
graph{(y-(2x^2+1)/x)(y+x+4)((y+3)^2+(x+1)^2-0.01)=0 [-10, 10, -5, 5]}