1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find the equation of a parabola
Loading...