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)#?

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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]}