# What is the equation of the normal line of f(x)=x^3 + 3x^2 + 7x - 1  at x=-1 ?

Aug 29, 2016

$y = \frac{x}{4} + \frac{23}{4}$

#### Explanation:

$f \left(x\right) = {x}^{3} + 3 {x}^{2} + 7 x - 1$
The gradient function is the first derivative
$f ' \left(x\right) = 3 {x}^{2} + 6 x + 7$

So the gradient when X=-1 is 3-6+7=4
The gradient of the normal , perpendicular, to the tangent is $- \frac{1}{4}$

If you are not sure about this draw a line with gradient 4 on squared paper and draw the perpendicular.

So the normal is $y = - \frac{1}{4} x + c$

But this line goes through the point (-1,y)
From original equation when X=-1 y=-1+3-7-1=6

So 6=$- \frac{1}{4} \cdot - 1 + c$
$C = \frac{23}{4}$