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

Feb 11, 2015

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

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.

$\frac{d}{\mathrm{dx}} y = \frac{d}{\mathrm{dx}} \left(16 {x}^{-} 1 - {x}^{2}\right)$
$\frac{d}{\mathrm{dx}} y = - 16 {x}^{-} 2 - 2 x$

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.

$0 = - 16 {x}^{-} 2 - 2 x$
$2 x = - \frac{16}{x} ^ 2$
$2 {x}^{3} = - 16$
${x}^{3} = - 8$
$x = - 2$

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.

$y = 16 {\left(- 2\right)}^{-} 1 - {\left(- 2\right)}^{2} = - 8 - 4 = - 12$

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

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]}