What is the equation of the normal line of $f(x)=-x^4+2x^3-12x^2-13x+3$ at $x=-1$ ?

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Answer 1 · 2016-02-08T17:54:45

$y=-x/21+62/21$

Let $y=-x^4+2x^3-12x^2-13x+3$

$dy/dx=-4x^3+6x^2-24x-13=m$

At $x=-1$ we get:

$m=4+6+24-13=21$

If the gradient of the normal line is $m'$ then:

$m'.m=-1$

$:.m'=-1/m=-1/21$

The value of $y$ at $x=-1$ becomes:

$y=1-2-12+13+3=3$

The equation of the normal is of the form:

$y=m'x+c$

$:.3=-1/21.(-1)+c$

$:.c=62/21$

So the equation of the normal is:

$y=-x/21+62/31$

This is shown here:

graph{(y+x^4-2x^3+12x^2+13x-3)(y+x/10-62/21)=0 [-20, 20, -10, 10]}

What is the equation of the normal line of f(x)=-x^4+2x^3-12x^2-13x+3f(x)=-x^4+2x^3-12x^2-13x+3 at x=-1x=-1?