How do you find the points where the tangent line is horizontal given y=16x^-1-x^2?

1 Answer
Kuba Dolecki
Feb 11, 2015

The point at which the tangent line is horizontal is (-2, -12).

To find the points at which the tangent line is horizontal, we have to find where the slope of the function is 0 because a horizontal line's slope is 0.

d/dxy = d/dx(16x^-1 - x^2)
d/dxy = -16x^-2 - 2x

That's your derivative. Now set it equal to 0 and solve for x to find the x values at which the tangent line is horizontal to given function.

0 = -16x^-2 - 2x
2x = -16/x^2
2x^3 = -16
x^3 = -8
x = -2

We now know that the tangent line is horizontal when x = -2

Now plug in -2 for x in the original function to find the y value of the point we're looking for.

y = 16(-2)^-1 - (-2)^2 = -8 - 4 = -12

The point at which the tangent line is horizontal is (-2, -12).

You can confirm this by graphing the function and checking if the tangent line at the point would be horizontal:

graph{(16x^(-1)) - (x^2) [-32.13, 23, -21.36, 6.24]}

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