# 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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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.

`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.