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Simple Function (x & y intercepts)

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    For the function f(x) = 1/4 (x=2)^2 - 4, i'm confused with regards to finding x & y intercepts.

    2. Relevant equations

    3. The attempt at a solution

    You can work them out from the function in its current form, i think? For y-int, let x=0, which leaves (0,-3).

    For x-nt, i done as follows;

    1/4 (x+2)^2 = 4
    (x+2)^2 = 16
    x+2 = [tex]\pm[/tex] 4

    x=2, x= -6

    But then also one could use the quadratic formula, working from

    1/4 (x^2 +4x +4) -4
    1/4x^2 + x -3

    Which gives intercepts 3/2, and -1/2

    And also f(4x)= (x+2)^2 - 16
    = x^2 + 4x - 12

    4x=2, 4x= -6

    x=1/2, x= -3/2

    So, 3 different sets. What am i missing? I mean, i'm certain that the vertex lies at (-2,-4), so x=2, x=-6 must be the correct intercepts - but why are the others wrong?

  2. jcsd
  3. Oct 20, 2008 #2


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    Science Advisor

    No, it doesn't. here "a" is 1/4, "b" is 1, and "c"=-3 so
    x= (-1+ sqrt(1-(4(1/4)(-3)/(2*1/4)= (-1+ sqrt(4))/(1/2)= 1/(1/2)= 2 and
    x= (-1- sqrt(1-(4(1/4)(-3)/(2*1/4)= (-1- sqrt(4))/(1/2)= -3/(1/2)= -6.

    Did you forget the "a" in the "2a" in the denominator?

    No, f(4x)= (1/4)(4x- 2)^2-4. Did you mean 4f(x)? If f(x)= 0, then 4f(x)= 0 for exactly the same x.

    which gives the roots you got initially.

    But it was NOT 4x, it was 4f(x)

    You weren't "missing" anything- just did the algebra wrong.
  4. Oct 20, 2008 #3

    Ohhh. Damn. Yes, sorry, and thanks. I often do that actually, */2 instead of */2a using the quadratic formula. I must stop.

    Thanks, Hallsofivy.
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