Question

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

Answer 1

`y=x/4+23/4`

Explanation:

`f(x)=x ^3+3x^2+7x-1`
The gradient function is the first derivative
`f'(x)=3x^2+6x+7`

So the gradient when X=-1 is 3-6+7=4
The gradient of the normal , perpendicular, to the tangent is `-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=-1/4x+c`

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

So 6=`-1/4*-1+c`
`C=23/4`