Rewrite the quadratic equation y = 2x^2 + 10x + 9 into Standard Form to determine the Vertex.

Vertex:

Does it open up or down:

Is it taller or shorter than y = x^2:

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`y=2x^2+10x+9`

`=2(x^2+5x+9/2)`

`=2(x^2+2xx x xx (5/2)+(5/2)^2-(5/2)^2+9/2)`

`=2((x+5/2)^2-25/4+9/2)`

`=2((x+5/2)^2-7/4)`

`=2(x+5/2)^2-7/2`

`y+7/2=2(x+5/2)^2`

`(1/2)(y+7/2)=(x+5/2)^2`

So vertex

`y+7/2= 0`

`=> y=-7/2`

`x+5/2=0`

`=>x=-5/2`

Vertex is `(-5/2,-7/2)`

It will open up.

Red is graph for `y=2x^2+10x+9`

Green is graph for `y=x^2`

Thus it taller than y = x^2 (since red is taller than green).

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