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Things i forgot how to do/unsure about. Porabola stuff

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data
    1) find the vertex, 2) find the x intercepts in terms of a, h, and k, 3) explain how values of a, h, and k affect the graph

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

    3. The attempt at a solution
    1) (H,K)
    2) aaaahhh i don't know... i just forgot.
    3) A determines how wide it is, with a bigger a making a skinnier parabola, a smaller A making a wider parabola, and a negative a making it go upside down. H moves the vertex along the x axis, and K moves it along the Y axis.

    is the ones i answered right? also. i would really appreciate a hint as to how to find out 2.

    thanks guys
  2. jcsd
  3. Jan 22, 2008 #2
    2. In order for you to be on the x-axis, set y=0. So from there, just solve for x. Whenever a problem says solve "x" in terms of a, b, c ... etc., they just want you to have it look like x = a, b, c.

    1 & 3 are good.
  4. Jan 22, 2008 #3
    ooo... hm lets check it out.
  5. Jan 22, 2008 #4
    im getting that

    A(x-h)^2 = [tex]\sqrt{-k}[/tex]

    gives you one and if you subtract -K from this answer, you get the other x axis.
  6. Jan 22, 2008 #5
    x = [tex]\frac{\sqrt{-k}}{A} \pm AH[/tex]

    is that right? i am so braindead.
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