Question

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

Answer 1

`y=-1/11x+21/11`

Explanation:

Given -

`y=2x^4+4x^3-2x^2-3x+3`

The first derivative gives the slope of the curve at any given point.

`dy/dx=8x^3+12x^2_4x-3`

Slope of the curve exactly at `x=-1`

`m_1=8(-1)^3+12(-1)^2-4(-1)+3`

`m_1=-8+12+4+3=11`

This is the slope of the tangent also.

Normal cuts the tangent vertically.

So its slope is

`m_2=-1/11` [Since; `m_1 xx m_2=-1`]

The normal passes through `x=-1`

Find the Y-coordinate

`y=2(-1)^4+4(-1)^3-2(-1)^2-3(-1)+3`
`y=2-4-2+3+3=2`

The Normal passes through the point `(-1, 2)`

Then find the equation of the Normal

`mx+c=y`

`(-1/11)(-1)+c=2`
`1/11+c=2`
`c=2-1/11=21/11`

The equation of the Normal is-

`y=-1/11x+21/11`