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I can really use a hand on finding vertex

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    find the slandered equation of a parabola that has a vertex axis that satisfies the given condition.
    X intercepts -3 and 5
    Y coordinate 4

    2. Relevant equations
    A(x-h)^2+k=y


    3. The attempt at a solution
    -4(x-1)^2+4
     
  2. jcsd
  3. Jan 30, 2016 #2

    SteamKing

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    I don't know what a "slandered" equation is. Do you mean find the "standard" equation of a parabola?
     
  4. Jan 30, 2016 #3

    Merlin3189

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    If you have 3 points, you can make 3 equations, then solve to find the three constants in the parabola equation.
     
  5. Jan 30, 2016 #4
    yes I do
     
  6. Jan 30, 2016 #5

    SteamKing

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    I don't know what you've done here. Your attempt is not an equation. If you know the x-intercepts of the parabola, can't you deduce the standard form of the parabola's equation from that?
     
  7. Jan 30, 2016 #6

    SammyS

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    Hello AstralTao. Welcome to PF !

    I have corrected what you posted in the above quote by making my own assumptions.

    In your attempt at a solution, what you have given is NOT an equation. I assume you meant y = -4(x-1)2 + 4 . This does not give the correct x-intercepts.

    You need to show us how you arrived at that answer, if we are going to be able to help you.
     
  8. Jan 30, 2016 #7

    ehild

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    Do you really mean that the vertex axis has x intercepts?
     
  9. Jan 31, 2016 #8

    Merlin3189

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    I had taken the original post to mean the parabola with x intercepts -3 and 5, and y intercept 4.
    When Sammy said "y coord of vertex" I wondered if it meant that the maximum of the parabola had y coord = 4.
    This seems to make the arithmetic simpler (than y intercept = 4) especially using that general form of the parabola equation, so maybe that is what was meant.
    I can't remember hearing the term "vertex" used in this context, but it would be a reasonable one to describe the maximum point.
    What a "vertex axis" would be, I don't know. Maybe the vertical line of symmetry through the maximum. But then I'd expect to be told its x value, not its y value. It would obviously have one x intercept at (1,0) but no other intercepts, unless we start talking about the vertex axis intercepting the parabola.

    Were I answering the question, I'd just state clearly what I took the question to mean and solve that.
     
  10. Jan 31, 2016 #9

    HallsofIvy

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    If a parabola had x-intercepts a and b, then its equation is of the form y= A(x- a)(x- b) for some number A. You are given the x-intercepts so all you need to do is find the correct value of A. You say "Y coordinate 4"? The y coordinate of what point? Is that the y-intercept, where x= 0, or the y-coordinate of the vertex?
     
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