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

**Solution:**

Given, y= 16x^{-1} - x^{2}

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

dy/dx = d(16x^{-1} - x^{2})

dy/dx = -16x^{-2} - 2x

To find the slope of the horizontal line, set dy/dx to zero.

-16x^{-2} - 2x = 0

It can be written as

2x = -16/x^{2}

2x^{3} = -16

Dividing both sides by 2

x^{3} = -8

Taking cube root,

x = -2

This implies that the tangent line is horizontal at x = -2

Put the value of x in the given function,

y =16(-2)^{-1} - (-2)^{2}

y = -8 - 4

y = -12

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

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

**Summary:**

The points where the tangent line is horizontal for the given function y= 16x^{-1} - x^{2} is (-2, -12).

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