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=-1/13x+53/13`

Explanation:

Given -

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

The first derivative gives the slope at any given point

`dy/dx=8x^3+12x^2-4x-3`

At `x=1` the slope of the curve is -

`m_1=8(1^3)+12(1^2)-4(1)-3`
`m_1=8+12-4-3=13`

This is the slope of the tangent drawn to the point `x=1` on the curve.

The y-coordinate at `x=1`is

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

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

The normal and the tangent are passing through the point `(1, 4)`

The normal cuts this tangent vertically. Hence, its slope must be

`m_2=-1/13`

[You must know the product of the slopes of the two vertical lines is `m_1 xx m_2=-1` in our case `13 xx - 1/13 =-1`

The equation of the normal is -

`-1/13(1) +c=4`

`c=4+1/13=(52+1)/13=53/13`

`y=-1/13x+53/13`

Answer 2

`x+13y=53` or `y=-x/13+53/13`

Explanation:

`f(x)=2x^4+4x^3-2x^2-3x+3`

To find the equation to the normal First step is to find the slope.

The first derivative of a curve at a particular point is the slope of the
tangent at that point.

Use this idea let us first find the slope of the tangent

`f'(x)=8x^3+12x^2-4x-3`

`f'(1)=8+12-4-3=13`

The slope of the tangent to the given curve at x=1 is 13

The product of the slopes of the tangent and normal would be -1 .
so the slope of the normal is `-1/13.`

we need to find f(x) at `x=1, f(1)=2+4-2-3+3=4`
we have slope is `-1/13` and the point is (1,1).

We have `m=-1/13` and `(x1,y1)rarr(1,4)`

`y-4=(-1/13)(x-1)`
`13(y-4)=(-1)(x-1)`
`13y-52=-x+53`
`x+13y=53`