Question

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

Answer 1

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]}