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anonymous

  • one year ago

HELP?? WILL MEDAL, AND FAN Find the x-intercepts of the parabola with vertex (-1,-108) and y-intercept (0,-105). Write your answer in this form: (x1,y1),(x2,y2). If necessary, round to the nearest hundredth.

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  1. anonymous
    • one year ago
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    i have the answer so far, can anyone just help me get the points out of y=3x^2+2x-107???

  2. anonymous
    • one year ago
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    long time since i done these...

  3. anonymous
    • one year ago
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    :(((

  4. anonymous
    • one year ago
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    y=3x^2+6x-105

  5. anonymous
    • one year ago
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    got it

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  6. anonymous
    • one year ago
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    general form is y = ax^2 + bx + c when x = 0 , y = -105 so c = -105

  7. anonymous
    • one year ago
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    x coordinate of the vertex = -b / 2a = 1 the y coordinate of the vertex = -108 so -108 = a*1^2 + b - 105

  8. anonymous
    • one year ago
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    so b = -2a and -108 = a + b - 105 a + b = -3 substitute b = -2a in the above: a - 2a = -3 -a = -3 a = 3 now finding b: b = -3 -3 = -6 so equation is 3x^2 - 6x - 105 = 0 solve to find the x intercepts: (3x - 21)(x + 5 ) = 0 x = 7 , -5 x intercepts are 7 and -5

  9. anonymous
    • one year ago
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    you, but reverse the signs and its correct

  10. anonymous
    • one year ago
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    (5,0),(-7,0)

  11. anonymous
    • one year ago
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    ik

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