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Math Help - Slope of a graph

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question Slope of a graph

    i tried it, but i got the answer of 6, not 2.

    g(x) = x^2-4x , (3,-3)

    use the limit process to find the slope of the graph of the function at the specified point.

    i think my math is totally off
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  2. #2
    MHF Contributor red_dog's Avatar
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    The slope is

    g'(3)=\lim_{x\to 3}\frac{g(x)-g(3)}{x-3}=\lim_{x\to 3}\frac{x^2-4x+3}{x-3}=

    =\lim_{x\to 3}\frac{(x-1)(x-3)}{x-3}=\lim_{x\to 3}(x-1)=2
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  3. #3
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    i tried it, but i got the answer of 6, not 2.

    g(x) = x^2-4x , (3,-3)

    use the limit process to find the slope of the graph of the function at the specified point.

    i think my math is totally off
    If you have a functiony = f(x) the slope at x = a is calculated by:

    y'=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}

    With your example you have:

    g'(3)=\lim_{h\to 0}\dfrac{((3+h)^2-4(3+h))-(3^2-4\cdot 3)}{h}

    Expand the brackets, factor out h in the numerator, cancel h, and then calculate the value if h approaches zero:

    \lim_{h\to 0}\dfrac{((3+h)^2-4(3+h))-(3^2-4\cdot 3)}{h} =\lim_{h\to 0} \dfrac{9+6h+h^2-12-4h-9+12}{h}= \lim_{h\to 0}\dfrac{2h+h^2}{h}=\lim_{h\to 0}(2+h) = 2
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