# What is the equation of the line normal to f(x)= -1/x  at x=-3?

Mar 12, 2018

$\textcolor{b l u e}{y = - 9 x - \frac{82}{3}}$

#### Explanation:

The line that is normal to $f \left(x\right) = - \frac{1}{x}$ at $x = - 3$ is perpendicular to the tangent line at $x = - 3$.

We first need to find the gradient of the tangent line at $x = - 3$. We now find the derivative of $f \left(x\right) = - \frac{1}{x}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(- \frac{1}{x}\right) = \frac{1}{x} ^ 2$

Plugging in $x = - 3$:

$\frac{1}{- 3} ^ 2 = \frac{1}{9}$

We know that if two lines are perpendicular, then the product of their gradients is $- 1$

Let gradient of the normal be $\boldsymbol{m}$, then:

$m \cdot \frac{1}{9} = - 1 \implies m = - 9$

The normal passes through the point with x coordinate $- 3$. We now need a corresponding $y$ coordinate. Plugging in $x = - 3$ into $f \left(x\right) = - \frac{1}{x}$:

$- \frac{1}{- 3} = \frac{1}{3}$

Using point slope form of a line:

$\left({y}_{2} - {y}_{1}\right) = m \left({x}_{2} - {x}_{1}\right)$

$y - \left(\frac{1}{3}\right) = - 9 \left(x - \left(- 3\right)\right)$

$\textcolor{b l u e}{y = - 9 x - \frac{82}{3}}$

Graph: 