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

##### 1 Answer

Aug 27, 2016

#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

#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

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

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#