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

The y-intercept will occur at

The function's derivative can be found by the power rule, which states that

The slope of the tangent is given by evaluating your point

The slope of the tangent is

Since the reaction passes through

Here is a graphical depiction of the problem. The graph in

Hopefully this helps!

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To find the normal to the tangent line at the y-intercept of the curve (y = \frac{1}{3}x^3 - 4x + 2), you first need to find the derivative of the function, then find the slope of the tangent line at the y-intercept, and finally determine the negative reciprocal of this slope to find the slope of the normal line.

The derivative of the function (y = \frac{1}{3}x^3 - 4x + 2) is (y' = x^2 - 4).

At the y-intercept, (x = 0), so the y-intercept is (y = \frac{1}{3}(0)^3 - 4(0) + 2 = 2).

Substituting (x = 0) into the derivative, we find that the slope of the tangent line at the y-intercept is (y' = (0)^2 - 4 = -4).

The negative reciprocal of -4 is (\frac{-1}{-4} = \frac{1}{4}), so the slope of the normal line is (m = \frac{1}{4}).

Therefore, the normal to the tangent line at the y-intercept of the curve (y = \frac{1}{3}x^3 - 4x + 2) has a slope of (\frac{1}{4}).

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