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Math Help - equation of the tangent to the curve

  1. #1
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    equation of the tangent to the curve

    hi guys im having trouble figuring this out.

    Find the equation of the tangent to the curve

    <br />
y = \frac {x^2} {x-1}<br />

    when

    <br />
x = -1<br />

    thank u!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by jvignacio View Post
    hi guys im having trouble figuring this out.

    Find the equation of the tangent to the curve

    <br />
y = \frac {x^2} {x-1}<br />

    when

    <br />
x = -1<br />

    thank u!
    Let f(x)=\frac {x^2} {x-1}

    The equation of the tangent to the curve of f at a is y=f(a)+(x-a)f'(a). In this case, a=-1 hence what you've to do is simply computing f(-1) and f'(-1).

    To find the derivative of f, you can use the quotient rule : \left(\frac{u}{v}\right)'=\frac{u'v-v'u}{v^2}

    Does it help ?
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  3. #3
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    yes we did that but what we are confused about is if we use x = -1 for both points?

    because we will input -1 into the original formula, then the first point will be (-1, y(original))

    then we get the derivative of the original equation then input x into that equation which will give us (-1,y(derivative of original)) ??????

    but then if we find m, the denominator will be equal to ZERO which will give us an undefined function.

    ??
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  4. #4
    Newbie SuumEorum's Avatar
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    You find the equation of the tangent by using y-y_{1}=m(x-x_{1}) where x_{1} and y_{1} are coordinates for a point on the tangent, in your case, (-1,f(a)). m is found by substituting -1 into y', which you get by using the quotient rule.
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  5. #5
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    Yesss i got it! thank u
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