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How to find function with given roots and point

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the quadratic function which has x-intercepts -2 and 2 and passes through the point (0,8).

    I am struggling with solving this one. I tried to substitute values into ax^2+bx+c=y but there are too many unknowns.

    Can anyone give some tips how to solve it?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 31, 2012 #2

    micromass

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    Substituting values in [itex]y=ax^2+bx+c[/itex] is a good approach. Can you tell us what you get if you do this?
     
  4. Jan 31, 2012 #3
    Well, I know c = 8, but I still don't know the values of a and b.

    ax^2+bx+8=y.
     
  5. Jan 31, 2012 #4

    micromass

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    -2 and 2 are x-intercepts, what does that mean?? What points lie on the parabola??
     
  6. Jan 31, 2012 #5
    It means that they cross the x axis. And where they cross y=0... But then I get this:

    a(2)^2+2b+8=0

    I still have two unknowns.
     
  7. Jan 31, 2012 #6

    micromass

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    Yes, so 2 is an x-intercept thus

    4a+2b+8=0

    Also -2 is an x-intercept, thus: ...


    You'll end up with a system of 2 equations and 2 unknowns.
     
  8. Jan 31, 2012 #7
    Ok, got it:

    -2x^2+8=y

    Thanks.
     
  9. Jan 31, 2012 #8

    LCKurtz

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    You also might have noticed that, given the roots -2 and 2 your equation must have been ##y = c(x-2)(x+2) = c(x^2-4)## which would have had only one unknown.
     
  10. Jan 31, 2012 #9
    And what if the point which the function crosses is not y-intercept, eg (4,18)?
     
  11. Jan 31, 2012 #10

    LCKurtz

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    Are you addressing that question to me? What difference does that make? Put in x = 4, y = 18 and solve for c.
     
  12. Jan 31, 2012 #11
    But then I don't have b and a.
     
    Last edited: Jan 31, 2012
  13. Jan 31, 2012 #12

    LCKurtz

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    Huh?? If you put x=4, y = 18 into y = c(x2-4) you get 18 = c(16-4).
     
  14. Jan 31, 2012 #13
    Sorry, I meant b and a and was referring to the equation in form ax^2+bx+c=y. How would I go about using this form to solve the problem?

    Also, could you please explain in more detail this:

    y=c(x−2)(x+2)

    I think I am seeing this form for the first time, but I might be wrong.
     
  15. Jan 31, 2012 #14

    LCKurtz

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    Presumably you know the factor theorem: If r is a root of a polynomial, then (x-r) is a factor. So your two given roots imply that (x-2) and (x+2) are factors. Two linear factors make a quadratic and the only question is a possible multiplier, or coefficient of x2. That is you you can deduce the form of the polynomial must be:

    y=c(x−2)(x+2).

    Then putting in your other point determines c. It is equivalent to how you solved the problem, but a little easier.
     
  16. Jan 31, 2012 #15
    Yes, I know the factor theorem. But I cannot understand how you can deduce y=c(x-2)(x+2). Do you suggest that c is coefficient of x^2 (of course it is to)?
     
  17. Jan 31, 2012 #16

    eumyang

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    You are given that 2 and -2 are the x-intercepts. That means that
    (x - 2) and (x - (-2)) = (x + 2)
    are factors of this quadratic.

    So y has to equal
    y = a(x - 2)(x + 2)
    (I don't like using c -- to be consistent with the OP I'll use a)
    Plug in the point (0, 8) and you'll have a.
     
  18. Jan 31, 2012 #17
    I understand that part. But how did the c get before brackets?
     
  19. Jan 31, 2012 #18

    eumyang

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    Say that 3 and 5 are the zeros of another quadratic. Then, without any further information, you could say that the quadratic is
    y = (x - 3)(x - 5) = x2 - 8x + 15.

    But that's not the only quadratic that would work. This one would also work:
    y = 5x2 - 40x + 75
    If you find the zeros of this quadratic, you will find that they will also be 3 and 5. This quadratic happens to be 5 times the first quadratic:
    5x2 - 40x + 75 = 5(x2 - 8x + 15)

    In general, given the zeros of a quadratic p and q (and nothing else), the quadratic would be
    y = a(x - p)(x - q)

    However, in the OP, a point is also given, which means that we need to find a specific value of a that would allow the quadratic to go through that point.


    P.S. You realize that the c in y = c(x - 2)(x + 2) is not the same as the c in y = ax2 + bx + c, do you?
     
  20. Jan 31, 2012 #19
    No. I actually realized that just now :). I thought you meant "c" as y-intercept. But now that you explained that it is rather "a", then I understand the logic.

    Thanks for detailed explanation.
     
  21. Jan 31, 2012 #20

    LCKurtz

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    Good job of catching that eumyang. I hadn't caught that his difficulty was with the fact I had used "c" instead of "a" in my explanation.
     
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