# What is the equation of the line normal to f(x)=x^2+8x - 1 at x=5?

Feb 19, 2016

$y = - \frac{x}{18} + \frac{1157}{18}$

#### Explanation:

Let $y = {x}^{2} + 8 x - 1$

$y ' = 2 x + 8$

This is equal to the gradient of the tangent line $m$.

At $x = 5$:

$m = \left(2 \times 5\right) + 8 = 18$

If $m '$ is the gradient of the normal line then:

$m ' m = - 1$

$\therefore m ' = - \frac{1}{m} = - \frac{1}{18}$

and:

$y = {5}^{2} + \left(8 \times 5\right) - 1 = 25 + 40 - 1 = 64$

The equation of the normal line is of the form:

$y = m ' x + c$

$\therefore 64 = \left(- \frac{1}{18} \times 5\right) + c$

$\therefore c = 64 + \frac{5}{18} = \frac{1157}{18}$

So the equation of the normal line is:

$y = - \frac{x}{18} + \frac{1157}{18}$

graph{(y+x/18-1157/18)(x^2+8x-1-y)=0 [-180, 180, -90, 90]}