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

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Answer 1 · 2016-08-27T06:36:56

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

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-

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

Look at the graph

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