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

Question

2992 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T18:23:43

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

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

$y=a(x-r_1)(x-r_2)$

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 $x=(r_1+r_2)/2$ .

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 $x_v=(4+(-6))/2=-1$ .

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

$y_v=-2(-1-4)(-1+6)=50$

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

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)?