# How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of y=-2(x-4)(x+6)?

Apr 10, 2018

The parabola opens down and its vertex is at $\left(- 1 , 50\right)$.

#### Explanation:

You have a quadratic equation in factored form. Here is the general equation for a factored quadratic.

$y = a \left(x - {r}_{1}\right) \left(x - {r}_{2}\right)$

If $a > 0$, the parabola opens up. If $a < 0$, the parabola opens down.

The $x$-intercepts (roots) of the parabola are at $x = {r}_{1}$ and $x = {r}_{2}$.

The axis of symmetry is at $x = \frac{{r}_{1} + {r}_{2}}{2}$.

In this case, $a = - 2$, ${r}_{1} = 4$, and ${r}_{2} = - 6$.

Since $a = - 2 < 0$, the parabola open down.

The axis of symmetry is at ${x}_{v} = \frac{4 + \left(- 6\right)}{2} = - 1$.

The $y$-coordinate of the vertex, ${y}_{v}$ is the original equation evaluated at $x = - 1$.

${y}_{v} = - 2 \left(- 1 - 4\right) \left(- 1 + 6\right) = 50$

So the vertex is at $\left(- 1 , 50\right)$.

graph{-2(x-4)(x+6) [-10, 10, -60, 60]}