Question 80511: Directions: decide whether the graph of the equation opens up or down. Then find the coordinates of the vertex.as in ... y=3x^2 and/or y=2x^2-4x+3
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Directions: decide whether the graph of the equation opens up or down. Then find the coordinates of the vertex.as in ... y=3x^2 and/or y=2x^2-4x+3
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y= 3s^2 opens up.
Since a=3 and b=0 the x-coordinate of the vertex = -b/2a = -0/6=0
When x=0, y=0 so the vertex is at (0,0)
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y=2x^2-4x+3 opens up
Since a=2, and b=-4, x=-b/2a = -(-4/4)=1
When x=1, y=2-4+3=1 so the vertex is at (1,1)
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Cheers,
Stan H.
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