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- anonymous

without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.
y=2x^2+x-3
A. 2 points in common; vertex below x-axis
B. no points in common; vertex below x-axis
C. 1 point in common; vertex on x-axis
D. 2 points in common; vertex above x-axis

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- anonymous

- katieb

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- anonymous

Do you know how to tell whether the graph opens up or down?

- anonymous

if the coefficient of x² is positive the parabola opens up. If it's negative the parabola opens down

- anonymous

and to find the x-coordinate of the vertex use x = -b/(2a)

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- anonymous

so, the coefficient of x^2 is 2, right?

- anonymous

right

- anonymous

so which way does the parabola open?

- anonymous

up

- anonymous

ok. so now let's find the vertex. For a quadratic equation ax²+bx+c, the x-coordinate of the vertex is \[x=\frac{ -b }{ 2a }\]

- anonymous

i got -1/4=-0.25

- anonymous

right. Now plug -1/4 in for x in your original formula to find the y-coordinate.

- anonymous

i got -4.25? which answer is that?

- anonymous

I got -3.125

- anonymous

show me what you did, please

- anonymous

|dw:1433124251446:dw|

- anonymous

okay, so which answer is that? is the vertex below the x-axis?

- anonymous

right, below the x-axis

- anonymous

and if it open up, in how many places does it cross the x-axis?

- anonymous

i... i dunno.

- anonymous

i'm guessing two?

- anonymous

yes|dw:1433124522429:dw|

- anonymous

can you help me with one more? that'll be all I need help with. and thanks for helping me.

- anonymous

ok

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