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

Jun 24, 2016

$y = \frac{1}{5} x + \frac{32}{5}$

#### Explanation:

The normal line will intersect the curve at $\left(- 2 , f \left(- 2\right)\right)$. Since $f \left(- 2\right) = {\left(- 2\right)}^{2} - \left(- 2\right) = 6$, we know the normal line passes through the point $\left(2 , 6\right)$.

To find the slope of the normal line, first find the slope of the tangent line at that point by finding the value of the derivative at $x = - 2$. Then, since the normal line is perpendicular to the tangent line, take the opposite reciprocal of the slope of the tangent line.

Through the power rule, we see that $f ' \left(x\right) = 2 x - 1$. We then see that the slope of the tangent line at $x = - 2$ is $f ' \left(- 2\right) = 2 \left(- 2\right) - 1 = - 5$.

The slope of the normal line is then $- \frac{1}{- 5} = \frac{1}{5}$.

Using the point $\left(- 2 , 6\right)$ and slope of $\frac{1}{5}$, we can write the equation of the line from $y = m x + b$:

$6 = \frac{1}{5} \left(- 2\right) + b \text{ "=>" } b = \frac{32}{5}$

The normal line is:

$y = \frac{1}{5} x + \frac{32}{5}$

Graphed are $f \left(x\right)$ and the normal line:

graph{(y-x^2+x)(-y+1/5x+32/5)=0 [-15.61, 12.86, -1.46, 12.77]}