# What is the equation of the normal line of f(x)=-x^4+2x^3-12x^2-13x+3 at x=-1?

Feb 8, 2016

$y = - \frac{x}{21} + \frac{62}{21}$

#### Explanation:

Let $y = - {x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 13 x + 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 4 {x}^{3} + 6 {x}^{2} - 24 x - 13 = m$

At $x = - 1$ we get:

$m = 4 + 6 + 24 - 13 = 21$

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

$m ' . m = - 1$

$\therefore m ' = - \frac{1}{m} = - \frac{1}{21}$

The value of $y$ at $x = - 1$ becomes:

$y = 1 - 2 - 12 + 13 + 3 = 3$

The equation of the normal is of the form:

$y = m ' x + c$

$\therefore 3 = - \frac{1}{21.} \left(- 1\right) + c$

$\therefore c = \frac{62}{21}$

So the equation of the normal is:

$y = - \frac{x}{21} + \frac{62}{31}$

This is shown here:

graph{(y+x^4-2x^3+12x^2+13x-3)(y+x/10-62/21)=0 [-20, 20, -10, 10]}