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cos00155079

  • 3 years ago

can someone check my answer? find the vertex, just not sure about the negatives

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  1. cos00155079
    • 3 years ago
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  2. slaaibak
    • 3 years ago
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    correct

  3. slaaibak
    • 3 years ago
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    usually, to check which value is the x-coordinate of the vertex, just check what value of x will make the bracket zero. in this case it's (-4 + 4)^2 - 33 so x-coordinate = -4

  4. slaaibak
    • 3 years ago
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    oh, sorry, I never saw the equation was x = y^2 + 8y - 17 it's gonna be a bit different now, I'll post the solution now

  5. campbell_st
    • 3 years ago
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    the vertex is at (h, k) and the general form is y=(x -h )^2 + k but since the parabola x = y^2 + 8y - 17 |dw:1354919627767:dw|

  6. cos00155079
    • 3 years ago
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    yeah, well I have all the steps I'm just confused with the negatives

  7. slaaibak
    • 3 years ago
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    yeah, what cambpell said. so it will be (-33,-4)

  8. campbell_st
    • 3 years ago
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    no the parabola is concave right.... its x = y^2 + 8x - 17

  9. campbell_st
    • 3 years ago
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    the methodology is correct... its problem was identifying the type of parabola you have...

  10. cos00155079
    • 3 years ago
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    let me show u my steps ill upload a pic

  11. cos00155079
    • 3 years ago
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    I only need the vertex, but i dont have to graph it... wait a sec please

  12. campbell_st
    • 3 years ago
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    I've seen your method.... but the parabola is not the traditional concave up parabola...

  13. cos00155079
    • 3 years ago
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  14. cos00155079
    • 3 years ago
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    so it would be (-4,-33) ?

  15. cos00155079
    • 3 years ago
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    yes, @slaaibak but its (h,k) so (-4,-33) right?

  16. campbell_st
    • 3 years ago
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    the methodolgy is correct... here is the graph attached the error is in the general form that you have used. for parabola that are in the form x = ay^2 + by + c the vertex form is x = (y - k)^2 + h

  17. cos00155079
    • 3 years ago
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    ok got it! thanks a lot :)

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