# What is the equation of the normal line of f(x)= x^2 − 5x + 1 at x = 5?

Nov 16, 2015

$y = - \frac{1}{5} x + 2$

#### Explanation:

Find the slope of the tangent line at the point $\left(5 , 1\right)$ using the derivative. Then, write the equation of the line with a perpendicular slope at the point $\left(5 , 1\right)$.

Remember that $\frac{d}{\mathrm{dx}} \left[a {x}^{n}\right] = n a {x}^{n - 1}$, where $a$ is a constant.
$f ' \left(x\right) = 2 x - 5$
$f ' \left(5\right) = 5$

Therefore, if the slope of the tangent line at $x = 5$ is $5$, the perpendicular slope of the normal line is $- \frac{1}{5}$.

Write in point-slope form: $y - 1 = - \frac{1}{5} \left(x - 5\right)$

In slope-intercept form: $y = - \frac{1}{5} x + 2$