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anonymous

  • 5 years ago

completing the square to graph the parabola y=-2x^2-8x-4 I got stuck but here is what I have 20+y=2(x-4)^2

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  1. amistre64
    • 5 years ago
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    y+4 = -2x^2 -8x ; divide everything by -2 y-2 = x^2 +4x ; complete the square.. (4/2)^2..+4 to each side y-2+4 = x^2 +4x +4

  2. anonymous
    • 5 years ago
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    That's perfect. This is telling you that it's a translation in y down 20 units, and to the right in x 4 units.

  3. amistre64
    • 5 years ago
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    y+2 = (x+2)^2 1(y+2) = (x+2)^2

  4. amistre64
    • 5 years ago
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    forgot to divde y by -2 :)

  5. anonymous
    • 5 years ago
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    let me show you part of how my teacher showed me how to work it

  6. amistre64
    • 5 years ago
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    (y/-2) +2 = (x+2)^2 ; factor the -1/2 from the y (-1/2)(y-4) = (x+2)^2

  7. anonymous
    • 5 years ago
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    Actually though, your answer isn't right Marie.

  8. anonymous
    • 5 years ago
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    It's in the right form though

  9. amistre64
    • 5 years ago
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    the vertex is at (-2,4) and it opens downward....

  10. anonymous
    • 5 years ago
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    y+4=-2x^2-8x take the -8/2= -4 then square it which is 16 add 16 to both sides y+20=-2x^2-8x+16 now I take the right side and put it in paranthesis be fore it was added togather 20+y=-2(x-4)^2

  11. amistre64
    • 5 years ago
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    the usual look to this is: y = -2(x+2)^2 +4

  12. amistre64
    • 5 years ago
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    you cant complete the square when the first term constant is not = 1

  13. amistre64
    • 5 years ago
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    divide everything by -2 first

  14. anonymous
    • 5 years ago
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    \[y = -2x^2 -8x -4\] \[y = -2(x^2 + 4x + 2) \] \[y = -2(x^2 + 4x + 2 + (2 - 2))\] ^ Completing the square \[y = -2(x^2 + 4x + 4 -2)\] \[y = -2(x+2)^2 +4\] \[y -4 =-2(x+2)^2\]

  15. anonymous
    • 5 years ago
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    y+10=(x-4)^2

  16. anonymous
    • 5 years ago
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    Whoops, sorry. Didn't mean to jump the gun there

  17. amistre64
    • 5 years ago
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    its ok, thats what I got too :)

  18. anonymous
    • 5 years ago
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    You have to factor out the -2 first, then work inside to complete the square.

  19. anonymous
    • 5 years ago
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    is the point that i'm tryin to find to graph the parabola (4, -10)?

  20. amistre64
    • 5 years ago
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    no, its (-2,4) as the vertex

  21. anonymous
    • 5 years ago
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    ok thanks for your help

  22. anonymous
    • 5 years ago
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    I showed the work, so you can see where you went wrong.

  23. amistre64
    • 5 years ago
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    you have to factor out or divide out that -2 that is sitting in front of the x^2 before you can complete the square...

  24. anonymous
    • 5 years ago
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    Just be sure to factor off any coefficients from your \(x^2\) term

  25. anonymous
    • 5 years ago
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    Yeah.

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