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Math Help - Equation of tangent line to the graph

  1. #1
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    Equation of tangent line to the graph

    I need to find an equation of a tangent line to the graph of a function at the given point.


    Given:
    g(x)=(x+2)(\frac{x-5}{x+1}) at (0,-10)
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  2. #2
    MHF Contributor Amer's Avatar
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    derive

    g(x)=(x+2)\left(\frac{x-5}{x+1}\right)

    g'(x)=\left(\frac{x-5}{x-1}\right)+(x+2)\left(\frac{(x+1)-(x-5)}{(x+1)^2}\right)

    substitute 0 to get the slope

    g'(0)=17.....wright??

    now

    slope=\frac{y-y_0}{x-x_0}=\frac{y--10}{x-0}=17

    tangent...line....y+10=17x
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  3. #3
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    Quote Originally Posted by Amer View Post
    derive

    g(x)=(x+2)\left(\frac{x-5}{x+1}\right)

    g'(x)=\left(\frac{x-5}{x-1}\right)+(x+2)\left(\frac{(x+1)-(x-5)}{(x+1)^2}\right)

    substitute 0 to get the slope

    g'(0)=17.....wright??

    now

    slope=\frac{y-y_0}{x-x_0}=\frac{y--10}{x-0}=17

    tangent...line....y+10=17x

    Um isn't g'(x) = \frac{x-5}{x+1}+\frac{x+2}{x+1}-\frac{(x+2)(x-5)}{(x+1)^2}
    and thus:

    g'(0) = 7 ?

    who's calculation is right here?
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  4. #4
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    Quote Originally Posted by Amer View Post
    derive

    g(x)=(x+2)\left(\frac{x-5}{x+1}\right)

    g'(x)=\left(\frac{x-5}{x {\color{red}+} 1}\right)+(x+2)\left(\frac{(x+1)-(x-5)}{(x+1)^2}\right) Mr F says: Typo corrected (in red).

    substitute 0 to get the slope

    g'(0)=17.....wright?? Mr F says: So g'(0) = 7.

    [snip]

    Quote Originally Posted by hemi View Post
    Um isn't g'(x) = \frac{x-5}{x+1}+\frac{x+2}{x+1}-\frac{(x+2)(x-5)}{(x+1)^2}
    and thus:

    g'(0) = 7 ?

    who's calculation is right here?
    Amer made a small typo. It should be obvious how to fix it (even more so if you read the above) and therefore how to get the correct answer.
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  5. #5
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    yeh, it will simply become y+10=7x correct?
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  6. #6
    MHF Contributor Amer's Avatar
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    ohh
    sorry
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