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

Question

Ezekiel Alexander · Accepted Answer

Equation of normal is #x-12y+49=0#

We are seeking equation at #x=-1# on the curve #f(x)=-2x^3+4x^2+2x# i.e. at #(-1,f(-1))# i.e. #(-1,4)#

as #f(-1)=-2(-1)^3+4(-1)^2+2(-1)=2+4-2=4#

As slope of tangent at #f(-1)# is #f'(-1)# and as

#f'(x)=-6x^2+8x+2# and hence slope of tangent is #-6(-1)^2+8(-1)+2=-6-8+2=-12#

As normal is perpendicular to tangent, its slope is #(-1)/(-12)=1/12#

and equation of normal is #y-4=1/12(x+1)# or #x-12y+49=0#

graph{(x-12y+49)(y+2x^3-4x^2-2x)=0 [-9.71, 10.29, -1.28, 8.72]}

Emilia Aguilar · Answer

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

To find the slope of the tangent line, we take the derivative of the function f(x).

The derivative of f(x) = -2x^3 + 4x^2 + 2x is f'(x) = -6x^2 + 8x + 2.

To find the slope at x = -1, we substitute -1 into the derivative: f'(-1) = -6(-1)^2 + 8(-1) + 2 = -6 + (-8) + 2 = -12.

The slope of the tangent line at x = -1 is -12.

The negative reciprocal of -12 is 1/12.

Therefore, the equation of the normal line at x = -1 is y = (1/12)(x - (-1)) + f(-1).

Simplifying this equation, we have y = (1/12)(x + 1) + f(-1).

To find f(-1), we substitute -1 into the original function: f(-1) = -2(-1)^3 + 4(-1)^2 + 2(-1) = -2 + 4 - 2 = 0.

Therefore, the equation of the normal line at x = -1 is y = (1/12)(x + 1) + 0, which simplifies to y = (1/12)(x + 1).