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]}