# Rewrite the quadratic equation y = 2x^2 + 10x + 9 into Standard Form to determine the Vertex. Enter answers in Vertex, Up or Down, Taller or Shorter than y = x^2

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

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

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

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

So vertex of parabola is

`y+7/2=0`

`x+5/2=0`

`Solve above equations`

`x=-5/2,y=-7/2`

Thus vertex at (-5/2,-7/2)

Graph will move (open ) upword and defined only if `y>=-7/2`

This is larger than the `y=x^2` .

The second parabola has vertex at (0,0) ,which is above than the vertx of first parabola. We draw graph for first parabola in red color while green used for second.

1. In above answer vertx is incorrect.

**In fact vertex at (-5,-41) ,not at (-5,41).**

2. **In both cases graph will move upword if graph moved downward then vertex will repn. maximum not minimum.**

First case graph is defined only if y is greater or equals to -41.

while in second case it will be defined for all values of y ,greater or equals to zero.

3. Graph in first case is wider than the second case. So you can say first is larger than that of second.