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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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    Do you know how to tell whether the graph opens up or down?

  2. anonymous
    • one year ago
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    if the coefficient of x² is positive the parabola opens up. If it's negative the parabola opens down

  3. anonymous
    • one year ago
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    and to find the x-coordinate of the vertex use x = -b/(2a)

  4. anonymous
    • one year ago
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    so, the coefficient of x^2 is 2, right?

  5. anonymous
    • one year ago
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    right

  6. anonymous
    • one year ago
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    so which way does the parabola open?

  7. anonymous
    • one year ago
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    up

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

  9. anonymous
    • one year ago
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    i got -1/4=-0.25

  10. anonymous
    • one year ago
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    right. Now plug -1/4 in for x in your original formula to find the y-coordinate.

  11. anonymous
    • one year ago
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    i got -4.25? which answer is that?

  12. anonymous
    • one year ago
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    I got -3.125

  13. anonymous
    • one year ago
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    show me what you did, please

  14. anonymous
    • one year ago
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    |dw:1433124251446:dw|

  15. anonymous
    • one year ago
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    okay, so which answer is that? is the vertex below the x-axis?

  16. anonymous
    • one year ago
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    right, below the x-axis

  17. anonymous
    • one year ago
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    and if it open up, in how many places does it cross the x-axis?

  18. anonymous
    • one year ago
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    i... i dunno.

  19. anonymous
    • one year ago
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    i'm guessing two?

  20. anonymous
    • one year ago
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    yes|dw:1433124522429:dw|

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

  22. anonymous
    • one year ago
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    ok

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