# What is the equation of the normal line of f(x)=4x^4+8x^3-2x^2+x-3 at x=-1?

Apr 19, 2016

$y = - \frac{1}{13} x - \frac{131}{13}$

#### Explanation:

Let's start by writing down the equation of the normal line we are trying to find:

$y = {m}_{p} x + {b}_{p}$

The slope of a normal to a line with slope, $m$, is given by

${m}_{p} = - \frac{1}{m}$

If we can get slope of the tangent line to our function, we can use this expression to find the slope of the normal line. The slope of the tangent is just the first derivative of the function evaluated at the point of interest. The derivative is:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} f \left(x\right) = 16 {x}^{3} + 24 {x}^{2} - 4 x + 1$

the slope is then

$m = f ' \left(- 1\right) = - 16 + 24 + 4 + 1 = 13$

Therefore the slope of the normal line at is

${m}_{p} = - \frac{1}{13}$

Finally, to get the equation of the normal line we need to find the y-intercept, ${b}_{p}$, as well as the slope. To do this, we can plug in a known point which the normal line passes through. We know that it will pass through $y = f \left(x\right)$ at $x = - 1$, so we need to get the value of the original function at that point:

$y = f \left(- 1\right) = 4 - 8 - 2 - 1 - 3 = - 10$

Plugging this into the equation for the normal line we get:

$- 10 = - \frac{1}{13} \left(- 1\right) + {b}_{p}$

which gives

${b}_{p} = - \frac{131}{13}$

Finally, the equation of the normal line is

$y = - \frac{1}{13} x - \frac{131}{13}$

graph{((y+x/13+131/13) (y-(4x^4+8x^3-2x^2+x-3)))=0 [-20, 20, -19, 1]}