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Math Help - Sanity check on Q2 function

  1. #1
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    Sanity check on Q2 function

    Have two things bugging me -
    F(x)=(x-3)^2-5

    Wanted to find the x intercepts using the quadratic formula
    I wind up 6+- sqrt 20, how can I find x intercepts from that.
    I found the correct x intercepts using the 0=x^2-6x+4
    Then take the sqrt of 4 =2
    then (x-3)=2 (x-3) = -2
    x=5 x=1

    That matches the trace in my calculator, so I am sure this is correct but I just wish I could have found them using the quadratic method.

    The other thing that seems weird is the domain and range.
    my calculator shows all real numbers for domain, but the lowest number on the range was (-5, oo). Would like to see another way to prove the range.
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  2. #2
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    Those x-intercepts aren't right. You need the quadratic formula to find the x-intercepts for this one.

    To use the quadratic formula, you must have your quadratic in the form of : ax^{2} + bx + c = 0

    So:
    0 = (x-3)^{2} - 5 = x^{2} - 6x + 9 - 5
    0 = x^{2} - 6x + 4

    x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \: = \: \frac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(4)}}{2(1)} \: = \: \frac{6 \pm \sqrt{36 - 16}}{2} \: = \: \frac{6 \pm \sqrt{20}}{2}

    which are your x-intercepts.

    The domain of this function is basically all the x-values you can have which is any real number (there's not one number that will 'mess up'). For the range (the y-values the function can have), note that its vertex is at (3,-5) and opens upwards so it'll be impossible to have any y-value less than -5.
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  3. #3
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    Confused?

    So you are say one the X intercepts are 8.236 and the other is 3.764?
    That will not plot out with that vertex of (3,-5) -At least that is not how it is looking on my graph paper.
    Please explain, really confused as I can't see this visually, but I can with 1, 5 as the intercepts...
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  4. #4
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    f(x) = (x-3)^{2} - 5

    If x = 1 and x = 5 were REALLY your x-intercepts, then plugging them in should give you 0 right?

    f(1) = (1-3)^{2} - 5 = 4 - 5 = -1 \quad \text{Nope}
    f(5) = (5 - 3)^{2} - 5 = 4 - 5 = -1 \quad \text{Nope}

    And do you recognize the form of the parabola: y = a(x - b)^{2} + c? If you do, you should know that (b, c) represents your vertex.
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  5. #5
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    still don't get it...

    OK how do I plot the X intercepts and how does the function come out to Zero with them?
    What are the X intercepts?
    Sorry, I am probably a little slow...
    Thanks for the help
    Last edited by mathobust; May 2nd 2008 at 12:04 PM. Reason: it is not an equation
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  6. #6
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    The x-intercepts are the solutions to your quadratic equation which are: x = \frac{6 \pm \sqrt{20}}{2}

    You can simplify this even more if you want:
    = \frac{6 \pm \sqrt{4 \cdot 5}}{2} \: \: = \: \: \frac{6 \pm \sqrt{4}\sqrt{5}}{2} \: \: = \: \: \frac{6 \pm 2\sqrt{5}}{2} \: \: = \: \: 3 \pm \sqrt{5}

    So, estimate them:
    3 + \sqrt{5} \approx 5.236
    3 - \sqrt{5} \approx 0.764

    Then graph the points (5.236, 0) and (0,764, 0) and these will be your x-intercepts.
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  7. #7
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    thanks that helps

    I noticed that the points are fairly close to my 1 and 5...
    wonder if 1 and 5 will fly for the answer on my homework...?
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  8. #8
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    Sure just make the estimations. For example, 3 - \sqrt{5} \approx 0.764 is just a bit less than 1 so mark it as so.
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