Question

How do you find the points where the graph of the function `y= 1/ (x^2)` has horizontal tangents?

Answer 1

`+-oo`

Explanation:

`y=1/x^2 =x^(-2)`

`y` has horizontal tangents where `y'=0`

`y' =-2/x` (Power rule)

Thus `y` has horizontal tangents where `lim_"h->0" -2/h = +-oo`

From the graph below we can see that the `x`-axis is the horizontal asymtote to `y` and thus consistant with the result above.

graph{1/x^2 [-5.55, 5.55, -2.772, 2.78]}