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Math Help - Equation of a tangent line at a certain point.

  1. #1
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    Equation of a tangent line at a certain point.

    "Find the equation of the tangent line to the graph of y=(2x-5)/(x+1) at the point at which x=0."

    I thought I had started off doing this right, but I must've messed up somewhere.

    Don't I need to find the derivative of the function and then evaluate it at x=0?
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  2. #2
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    The equation of a tangent for this function is

     y-f(0) = f'(0)\times (x-0)

    Does this help?
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  3. #3
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    Well, I can form the tangent line once I have the values. But I think I'm coming out with the wrong values.

    Could you check my work below?

    To get the derivative, you must use the quotient rule.
    So you end up with f'(x)=((2(x+1))-(2x+5)(1))/(x+1)^2, which simplifies to f'(x)=((2x+1)-(2x+5))/(x+1)^2 and then f'(x)=-4/(x+1)^2. Is this right?
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  4. #4
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    You have expanded the function incorrectly.

    I get

     f'(x) = \frac{2(x+1)-1(2x-5)}{(x+1)^2}

     f'(x) = \frac{2x+2-2x+5)}{(x+1)^2}

     f'(x) = \frac{7}{(x+1)^2}
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