What is the equation of the normal line of #f(x)=(x-3)^3(x+2)# at #x=0#?

Question

Samantha Adkison · Accepted Answer

The equation of the normal line is #y=-1/27 x-54# Expanding the formula for #f# gives: #f(x)=(x^3-9x^2+27x-27)(x+2)# #=x^4-9x^3+27x^2-27x+2x^3-18x^2+54x-54# #=x^4-7x^3+9x^2+27x-54#. Therefore, #f'(x)=4x^3-21^2+18x+27#. Also note that #f(0)=-54# and #f'(0)=27#. The normal line at the point #(x,y)=(0,f(0))=(0,-54)# has slope equal to the negative reciprocal of #f'(0)=27#, so this slope equals #-1/27#. Since this normal line goes through the point #(0,-54)#, it follows that its equation is #y=-1/27x-54#.

Luke Alvarez · Answer

undefined To find the equation of the normal line of f(x)=(x-3)^3(x+2) at x=0, we need to find the slope of the tangent line at x=0 and then determine the negative reciprocal of that slope to find the slope of the normal line.

To find the slope of the tangent line at x=0, we take the derivative of f(x) and evaluate it at x=0.

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

Evaluating f'(x) at x=0, we get f'(0) = 3(0-3)^2(0+2) + (0-3)^3 = -27.

The slope of the tangent line at x=0 is -27.

The slope of the normal line is the negative reciprocal of the slope of the tangent line, so the slope of the normal line is 1/27.

Using the point-slope form of a line, we can write the equation of the normal line as y - f(0) = (1/27)(x - 0).

Simplifying, we get y - f(0) = (1/27)x.

Since x=0, we have y - f(0) = 0.

Therefore, the equation of the normal line of f(x)=(x-3)^3(x+2) at x=0 is y = f(0).