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Math Help - Function

  1. #1
    Junior Member
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    Function

    Could anyone help me on this:

    how to prove, that with all kind of a, b and c value, function's f(x) = (x-a)(x-b)-c^2 graph and Ox axis has at least one common point.

    What should I start at? Could you lead me to the right direction? thank you.
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  2. #2
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    Hello, Ellla!

    \text{Prove that, for all values of }a,b,c,\text{ the graph of the function:}

    . . f(x) \:=\: (x-a)(x-b)-c^2\,\text{ has at least one }x\text{-intercept.}

    An \,x-intercept occurs when f(x) = 0.


    We have: . (x-a)(x-b) - c^2 \;=\;0

    . . . . x^2 - (a+b)x + (ab-c^2) \:=\:0


    Quadratic Formula:

    . . x \;=\;\dfrac{(a+b) \pm\sqrt{(a+b)^2 - 4(ab-c^2)}}{2}

    . . . . =\;\dfrac{(a+b) \pm\sqrt{a^2 + 2ab + b^2 - 4ab + 4c^2}}{2}

    . . . . =\;\dfrac{(a+b)\pm\sqrt{a^2 - 2ab + b^2 + 4c^2}}{2}

    . . . . =\;\dfrac{(a+b) \pm\sqrt{(a-b)^2 + 4c^2}}{2} .[1]


    If a = b = c = 0, [1] has one value.
    . . The function has one \,x-intercept.

    For any other values of a,b,c, [1] has two values.
    . . The function has two x-intercepts.

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  3. #3
    MHF Contributor
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    One way it to see that f(a) = -c^2 <= 0. On the other hand, if x0 > max(|a|, |b|) + |c|, then f(x0) > 0. Therefore, f(x) must assume all intermediate values between f(a) and f(x0), including 0.

    Strictly speaking, this proof invokes the Intermediate value theorem from calculus, but it is one of the most obvious-looking theorems (not to say that its proof is trivial). Students of calculus are often puzzled why it has to be proved at all.
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  4. #4
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    Thank you.
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