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

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Answer 1 · 2018-04-10T18:23:43

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

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

$y = a (x - r_{1}) (x - r_{2})$ $y=a(x-r_1)(x-r_2)$

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

The $x$ $x$ -intercepts (roots) of the parabola are at $x = r_{1}$ $x=r_1$ and $x = r_{2}$ $x=r_2$ .

The axis of symmetry is at $x = \frac{r_{1} + r_{2}}{2}$ $x=(r_1+r_2)/2$ .

In this case, $a = - 2$ $a=-2$ , $r_{1} = 4$ $r_1=4$ , and $r_{2} = - 6$ $r_2=-6$ .

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

The axis of symmetry is at $x_{v} = \frac{4 + (- 6)}{2} = - 1$ $x_v=(4+(-6))/2=-1$ .

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

$y_{v} = - 2 (- 1 - 4) (- 1 + 6) = 50$ $y_v=-2(-1-4)(-1+6)=50$

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

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

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)y=-2(x-4)(x+6)?