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Homework Help: Find the equation of a parabola

  1. May 11, 2009 #1
    Find the equation of a parabola with the following characteristics:


    range Y <= 8
    x-coordinate of the turning point is -4
    y-intercept = -6


    I have tried to substitute all the information into y = a(x-p)^2 + q
    which gives me y = a(x+4)^2 + q and substituted the y-intercept into the equation and then subbed in q into the equation with the y-intercept subbed in as well but that just came up as 0 = 0

    Here is my working out

    Sub y-intercept into equation
    -6 = a(0 +4)^2 +q
    -6 = 16a + q
    q = -6 -16a

    Sub into equation (0,-6) and q
    -6 = a(0+4)^2 - 16a -6
    0 = 16a - 16a
    0 = 0
     
    Last edited: May 11, 2009
  2. jcsd
  3. May 11, 2009 #2

    HallsofIvy

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    Science Advisor

    Well, of course, solving for q using the y intercept and then putting the y-intercept into the equation will give you 0= 0! What did you expect?

    You haven't used all of the information. The fact that "range y<= 8" tells you that the vertex is at (-4, 8). That tells you both p and q. After you know p and q, the fact that the parabola passes through (0, -6) will give you a, which must be negative as the parabola opens downward.
     
  4. May 11, 2009 #3
    I'm assuming that "turning point" = "vertex"...

    If "range Y <= 8" then doesn't that imply that the y-coordinate of the vertex is 8? Can you take it from there?


    EDIT: Beaten to the punch by HallsofIvy! ;)

    01
     
  5. May 11, 2009 #4
    thanks for the help
     
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