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Homework Help: Parabolas; Domain and range

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    For the following questions, find the Domain and Range.

    Use either the graph or turning point method.
    2. Relevant equations
    1. y= -x+6x-8

    2. y= x^2-3x-5

    3. y= -x^2+4x+4

    And a question on the matter, when you complete the square, as in when I half the middle term square it plus take away etc... If before i did that, there is a negative, would i use that in my final answer?

    Very hard to write, check my attempts below, might be a bit more descriptive.

    3. The attempt at a solution
    I won't show the beginning of my working out, but what's near the end or it'll take forever.

    1. y= -x^2+6x-8
    -(x^2-6x+9)-9+8 = 0

    So what i was asking before was, the negative in front, does that affect the turning point? I wrote it as (3,-1), because the first is the opposite of what's inside and the other is what is outside the brackets exactly, but does that negative change what is inside the brackets that we use for the turning point? And also x is part of the real field right? because it is infinity.

    2. y= x^2-3x-5
    Turning point is = (1.5,-7.25)?

    3. y= -x^2+4x+4

    Turning point is = (2,-8) and again, does that negative in front affect anything?

    And also with this last one, Since it is y=-x^2+4x+4, does that not mean that the parabola will be negative? if it does, the Range would be that {y:y<-8} which means that there would be no X intercept, except for when you use the discriminant, it says that there are 2 answers?

    D = b^2-4ac
    = 4^2-(4x-1x4)
    = 16 + 16
    = 32

    If the discriminant > 0 doesn't that mean there are 2 x answers?

    Thanks for helping out!
  2. jcsd
  3. Sep 30, 2009 #2


    Staff: Mentor

    Where did y go? You are not solving the equation for the x-intercepts, just rewriting the original equation in a form suitable for finding the vertex. Here's what the above should look like:
    y= -x2+6x-8
    = -(x2 - 6x) - 8
    = -(x2 - 6x + 9) - 8 + 9 I'm adding 9 since I really added -9 earlier
    = (x - 3)2 + 1
    So y = (x - 3)2 + 1

    No. It affects only whether the parabola opens upward or downward.
    Because what is infinity? I don't understand your question here.
    = x2 - 3x + 9/4 - 5 - 9/4
    = (x - 3/2)2 - 29/4
    You have a mistake in the line above. What you really added inside the parentheses was -4, so to balance that, you need to add + 4, which makes your equation y = -(x - 2)2 + 8
    A parabola is neither negative nor positive. The negative coefficient on the x2 term determines that the parabola opens downward. A positive coefficient indicates that it will open upward.
  4. Sep 30, 2009 #3
    Thanks a heap mark you cleared up a lot for me, but to clarify:

    ok so when you said "I'm adding 9 since I really added -9 earlier" you meant that in:
    -(x2 - 6x + 9) - 8 + 9
    ^ Is actually a -9 because of the negative in front of the brackets?

    And in the transition between:

    = -(x2 - 6x + 9) - 8 + 9 I'm adding 9 since I really added -9 earlier
    = (x - 3)2 + 1

    What happened to the negative in front of the brackets? it just disappeared :S is it negated because you did the +9 at the end?

    Thanks for your help so far really helped me!
  5. Sep 30, 2009 #4


    Staff: Mentor

    Typo on my part. I neglected to bring it along. That last line should be
    = -(x - 3)2 + 1
  6. Sep 30, 2009 #5
    And so the TP of this would be (3,1)

    Thanks for all your help that's about it for now :)
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