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

Question

Elijah Abram · Accepted Answer

#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]}

Axel Alvarez · Answer

undefined To find the equation of the normal line at a given point on a curve, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line at x = -1, we first find the derivative of the function f(x) = -x^4 + 2x^3 - 12x^2 - 13x + 3.

Taking the derivative, we get f'(x) = -4x^3 + 6x^2 - 24x - 13.

Evaluating f'(-1), we have f'(-1) = -4(-1)^3 + 6(-1)^2 - 24(-1) - 13 = -4 + 6 + 24 - 13 = 13.

Therefore, the slope of the tangent line at x = -1 is 13.

The negative reciprocal of 13 is -1/13.

Hence, the equation of the normal line at x = -1 is y - f(-1) = (-1/13)(x - (-1)).

Simplifying this equation, we have y - f(-1) = (-1/13)(x + 1).