What is the normal to the tangent line at the y-intercept to #y = 1/3x^3 - 4x + 2#?

1 Answer
Dec 24, 2016

#y = 1/4x + 2#

Explanation:

The y-intercept will occur at #(0, 2)#:

#y = 1/3(0)^3 - 4(0) + 2 = 2#

The function's derivative can be found by the power rule, which states that #d/dx(x^n) = nx^(n - 1)#.

#y' = x^2 - 4#

The slope of the tangent is given by evaluating your point #x= a# into the derivative. The normal line is perpendicular to the tangent line, or the product of their two slopes equals #-1#.

The slope of the tangent is #m = 0^2 - 4 = 0 - 4 = -4#. Then the slope of the normal line is #1/4#.

Since the reaction passes through #(0, 2)#, the normal line has equation:

#y - y_1 = m(x - x_1)#

#y - 2 = 1/4(x - 0)#

#y - 2 = 1/4x#

#y = 1/4x + 2#

Here is a graphical depiction of the problem. The graph in #color(red)("red")# is the function #y = 1/3x^3 - 4x + 2#, the graph in #color(purple)("purple")# is the tangent line and the graph in #color(blue)("blue")# is the normal line.

enter image source here

Hopefully this helps!