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Parabolas Help

  1. Jan 8, 2008 #1
    1. The problem statement, all variables and given/known data

    How do you figure out the equation of a parabola by only knowing the vertex and ONE of the x intercepts

    Vertex: (0, -1920)
    X intercept: (96,0)

  2. jcsd
  3. Jan 8, 2008 #2


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    In general a parabola is of the form Ax^2 + Bx + C... but in your case one of the three constants happens to be zero. Which one?
  4. Jan 8, 2008 #3
    umm... :redface: .. not sure wht ur talkin about.. :confused:

    Last edited: Jan 8, 2008
  5. Jan 8, 2008 #4
    What is the general form of a quadratic equation?

    How is it written when we want to emphasize what the vertex is?
    Last edited: Jan 8, 2008
  6. Jan 8, 2008 #5
    well the general form of a parabola that is emphasized in vertex form is
  7. Jan 8, 2008 #6
    It's actually ...

    General form of a quadratic equation: [tex]y=ax^2+bx+c[/tex] & [tex]y=a(x-h)^2+k[/tex]

    Ok so, we know that h is the apex [tex]h=-\frac{b}{2a}[/tex] of our parabola. Since our vertex has points (0,-1920), we know that h=0 b/c of the fact it lies on the y-axis and that b=0 b/c in order for h to equal 0, b=0 ... if a equaled 0, then it would be undefined.

    So this reduces our standard quadratic equation to ...

    [tex]y=a(x-h)^2+k \rightarrow y=ax^2-1920[/tex]

    Further, since the apex of our parabola lies on the y-axis, we also know another zero, which is -96. Anyways, we can use either one since a parabola is symmetric with respect's to it's apex. Using our givens, [tex]P(\pm 96,0)[/tex], what we want to know now is the value for a ...

    Let's see your answer.
    Last edited: Jan 8, 2008
  8. Jan 9, 2008 #7
    k i made a huge mistake

    1. The problem statement, all variables and given/known data

    How do you figure out the equation of a parabola by only knowing the vertex and ONE OF THE POINTS ( not x intercept)

    Vertex: (0, 1920)
    X intercept: (960,0)

    its actaully POSITIVE 1960 and 960 not 96
  9. Jan 9, 2008 #8
    Well that point is an x-intercept b/c where exactly does P(960, 0) lie?

    Just follow what I told you in Post #6 and it's solved.
  10. Jan 10, 2008 #9


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    Now it makes no sense at all. Why would they say "ONE OF THE POINTS (not x intercept)" and then give you the x intercept?

    In any case, you are told that the vertex is at (0, 1920) so you know the equation is of the form y= a(x- 0)2+ 1920= ax2+ 1920. You only need to determine the single number, a. You also know that (960, 0) is a point on the parabola: that is, when x= 960, y= 0. Put those values of x and y into y= ax2+ 1920 and solve the equation for a.
  11. May 20, 2009 #10
    Ever heard of the line of symmetry?
    A parabola has one in the middle.

    1920 - 960 = 960
    1920 + 960 = 2780

    There! Two X-intercepts!
  12. May 20, 2009 #11
    0 = a921600 + 1920. Huh. It means a is -1/480. Wow. Amazing. And how come I am editting this after a year?
    Last edited: May 20, 2009
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