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Math Help - Solving Equation of Tangent Line to Curve

  1. #1
    salted
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    Solving Equation of Tangent Line to Curve

    I'm having trouble solving this problem. I have to find the equation of a line tangent to a curve at a specific point. I also have to use a certain technique.

    The curve is:

    f(x) = x^2+2x+3

    The point to be tangent to is:
    P = (-1, 6)

    I have to use these equations to solve.

    m(tan) = lim(x->c) (f(x) - f(c))/(x-c))

    and

    y - f(c) = m(tan)(x - c)

    I can get to the point where I have (x^2 - 2x) / (x+1). I'm not sure how to proceed. Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by salted
    I'm having trouble solving this problem. I have to find the equation of a line tangent to a curve at a specific point. I also have to use a certain technique.

    The curve is:

    f(x) = x^2+2x+3

    The point to be tangent to is:
    P = (-1, 6)
    As I understand this question you are asked to find the equation
    of the tangent to f(x)=x^2+2x+3 at the point P=(-1, 6).

    The problem with this is that P is not on the curve.

    RonL
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  3. #3
    MHF Contributor
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    Good point. Perhaps the OP meant the point (1,6)?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by salted
    I'm having trouble solving this problem. I have to find the equation of a line tangent to a curve at a specific point. I also have to use a certain technique.

    The curve is:

    f(x) = x^2+2x+3

    The point to be tangent to is:
    P = (-1, 6)

    I have to use these equations to solve.

    m(tan) = lim(x->c) (f(x) - f(c))/(x-c))

    and

    y - f(c) = m(tan)(x - c)

    I can get to the point where I have (x^2 - 2x) / (x+1). I'm not sure how to proceed. Thanks
    OK lets find the tangent at

    x=-1,

    f(-1)=1-2+3=2,

    so:

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

    hence:

    \lim_{x\rightarrow -1}\frac{f(x)-f(-1)}{x-(-1))}=\lim_{x\rightarrow -1}(x+1)=0

    Which is the horizontal line y=2

    -anyway this shows the technique of finding the slope you are required to
    use.

    RonL
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  5. #5
    Grand Panjandrum
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    OK now lets find the tangent at

    x=1,

    f(1)=1+2+3=6,

    so:

    \frac{f(x)-f(1)}{x-1}=\frac{x^2+2x-3}{x-1}=x+3

    hence:

    \lim_{x\rightarrow -1}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow 1}(x+3)=4

    So the tangent at P=(1,6) has slope 4, and
    applying your second equation gives the equation of the line as:

    y-f(c)=m(x-c)

    where c=1 and f(c)=6 so:

    y-6=4(x-1),

    or:

    y=4x+2

    RonL
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