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

Nov 7, 2016

$\left(0 , - 9\right)$

#### Explanation:

Observe that horizontal tangents occur at a min/max.

You don't need Calculus to answer this question:

$y = \left(x + 3\right) \left(x - 3\right)$

The function is a quadratic with roots given by:
$\left(x + 3\right) \left(x - 3\right) = 0 \implies x = \pm 3$
As the coefficient of ${x}^{2}$ is positive the quadratic will be $\cup$ 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 $x = 0 \implies y = \left(3\right) \left(- 3\right) = - 9$

So there is a horizontal tangent at $\left(0 , - 9\right)$

If you should want to use Calculus then:
$y = \left(x + 3\right) \left(x - 3\right) \implies y = {x}^{2} - 9$
$\therefore y ' = 2 x$
$y ' = 0 \implies 2 x = 0 \implies x = 0$ (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 $x = 0 \implies$ critical point is a minimum, as above.