Question

How do you find the points where the graph of the function `y=(x+3)(x-3)` has horizontal tangents?

Answer 1

`(0,-9)`

Explanation:

Observe that horizontal tangents occur at a min/max.

You don't need Calculus to answer this question:

`y = (x+3)(x-3)`

The function is a quadratic with roots given by:
`(x+3)(x-3) = 0 => x=+- 3`
As the coefficient of `x^2` is positive the quadratic will be `uu` shaped and so it will have a minimum, which by symmetry will occur at the midpoint of the roots, i.e when `x=0`

When `x=0=>y=(3)(-3)=-9`

So there is a horizontal tangent at `(0,-9)`

If you should want to use Calculus then:
`y = (x+3)(x-3) => y=x^2-9`
`:. y' = 2x`
`y' = 0=>2x =0 => x=0` (as above); so (0,-9) is a critical point

To establish the nature of the critical point we look at the second derivative:
`y'' = 2 > 0` when `x=0 =>` critical point is a minimum, as above.