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

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#

#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#