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Need pre calc help

  1. Mar 8, 2006 #1
    Find the standard and general form for the parabola with the vertex @ (2,-2) containing the point (3,0) and opening to the right.

    Who ever explains this I just want to know a hint, and please don't user mathematical terminology, I just want (in simple terms) how I can get to the next step.
  2. jcsd
  3. Mar 9, 2006 #2
    Well do you have any ideas on how to solve this?

    A parabola looks like [itex] y = x^2 [/itex] right?

    Now you are given a few conditions:
    1) vertex @ (2,-2)
    2) the parabola must contain the point (3,0)

    What are some things you can do to a parabola? How do you shift it along the x-axis (move it left and right)? How do you shift it along the y-axis (move it up and down)? How do you change how the parabola opens (opens up, or opens down)? How do you change how fast it increases or decreases?

    What happens when you change [itex] a [/itex] in the following function? Try both negative and positive.
    [tex] y = x^2 + a [/tex]

    What happens when you change [itex] b [/itex] in the following function? Again try negative and positive.
    [tex] y = (x+b)^2 +a [/tex]

    Play around with different values for [itex] a [/itex] and [itex] b [/itex] and see if you can get the parabola to open at your vertex.

    What happens when you change [itex] m [/itex] in the following function? Again try positive and negative.
    [tex] y = mx^2 [/tex]

    Now after you have [tex] a [/tex] and [tex] b [/tex] values try and find an appropriate [tex] m [/tex] value. So that you contain the point (3,0).

    Hint: Containing the point (3,0) means:
    [tex] x = 3 [/tex]
    [tex] y = 0 [/tex]

    Hint: The parabola will have the form:
    [tex] y = m(x+b)^2 +a [/tex]
  4. Mar 9, 2006 #3
    Since there is a root at (3,0), and a minimum at (2,-2), there must also be another point at (1,0) (by the symmetry of the situation, and then all you have to do is to write the equation as [itex] a(x - \alpha)(x - \beta) = y [/itex], where alpha and beta are the roots, a is just a constant (that you should be able to see by simply looking at the problem.

    Dam, beaten again.
  5. Mar 9, 2006 #4


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    No, that's not right and you are confusing things by starting off that way. A parabola can "look like" y= c(x-a)2+ b for any choice of a, b, c (except c= 0) or x= c(y-a)2+ b.

    Did you not notice the fact that this parabola "opens to the right"?
  6. Mar 9, 2006 #5
    Nope. I didn't notice that. If I did, I hope to god I wouldn't have said it would be in the form of the second hint (which really is not a helpful hint at all then). It was late... but not that late.

    I was introduced to the parabola by analyzing the function y=x^2 and seeing how to shift it with regards to whatever axis. A definition of a parabola that is almost as good as mine is here. :)

    Sorry about the confusion, hopefully it was late enough that the checkup posts were able to be read before I confused you to much.

    Also, a really helpful site that may be eaiser to read than Wolfram's is Purple Math.
  7. Mar 9, 2006 #6
    We know that a parabola looks like this:

    f(x)= x^2/a^2 + y^2/b^2 = 1 is that standard form

    Is this not correct?
  8. Mar 10, 2006 #7


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    No, that is not correct and we do NOT know that!!

    "x^2/a^2 + y^2/b^2 = 1" is the equation of an ellipse not a parabola!! PLEASE make sure of what you say before you try to "help"!

    That's two responses to this poor person that asserted things that are not true and surely are confusing to someone who did not know how to do the problem to start with!
  9. Mar 10, 2006 #8
    Ummm....HallsofIvy? konartist was the original poster. He was simply asking.

    Parabolas come in two formats:
    1) If they open up or down they are of the form: y = ax^2 + bx + c or y = d(x-e)^2+f.
    2) If they open to the right or left they are of the form: x = ay^2 + by + c or x = d(y-e)^2 + f.

    The axis of symmetry for 1) will be x = -b/(2a) (x = e for the second format) and the axis for 2) will be y = -b/(2a) (y = e for the second format.)

  10. Mar 10, 2006 #9


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    Oh! Okay, I shall calm down. No, konartist, what you gave was the equation of an ellipse not a parabola. (Although with a name like "konartist" you might just be pulling my leg!)
  11. Mar 10, 2006 #10
    OK, I understand.

    Sorry I knew that a standard form of a parabola is (x-h)^2=#(y-k)

    We're learning conics and it is hella confusing. I understand now, thank you.
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