anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Do you know how to tell whether the graph opens up or down?
anonymous
  • anonymous
if the coefficient of x² is positive the parabola opens up. If it's negative the parabola opens down
anonymous
  • anonymous
and to find the x-coordinate of the vertex use x = -b/(2a)

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anonymous
  • anonymous
so, the coefficient of x^2 is 2, right?
anonymous
  • anonymous
right
anonymous
  • anonymous
so which way does the parabola open?
anonymous
  • anonymous
up
anonymous
  • 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
  • anonymous
i got -1/4=-0.25
anonymous
  • anonymous
right. Now plug -1/4 in for x in your original formula to find the y-coordinate.
anonymous
  • anonymous
i got -4.25? which answer is that?
anonymous
  • anonymous
I got -3.125
anonymous
  • anonymous
show me what you did, please
anonymous
  • anonymous
|dw:1433124251446:dw|
anonymous
  • anonymous
okay, so which answer is that? is the vertex below the x-axis?
anonymous
  • anonymous
right, below the x-axis
anonymous
  • anonymous
and if it open up, in how many places does it cross the x-axis?
anonymous
  • anonymous
i... i dunno.
anonymous
  • anonymous
i'm guessing two?
anonymous
  • anonymous
yes|dw:1433124522429:dw|
anonymous
  • anonymous
can you help me with one more? that'll be all I need help with. and thanks for helping me.
anonymous
  • anonymous
ok

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