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

Question

Elijah Abram · Accepted Answer

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

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#

Isabella Ackerman · Answer

undefined To find the equation of the normal line of f(x) at x=-1, we need to determine the slope of the tangent line at that point. The slope of the tangent line is given by the derivative of the function at x=-1. Taking the derivative of f(x) with respect to x, we get f'(x) = 8x^3 + 12x^2 - 4x - 3. Evaluating f'(-1), we find f'(-1) = -5.

Since the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is 1/5.

Using the point-slope form of a line, we can write the equation of the normal line as y - f(-1) = (1/5)(x - (-1)). Simplifying this equation, we have y + 3 = (1/5)(x + 1).