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Math Help - Tangent to the curve

  1. #1
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    Tangent to the curve

    for what values of a and b will the parabola y =x^2 + ax + b be tangent to the curve y = x^3 at point (1,1)?
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  2. #2
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    Quote Originally Posted by mathamatics112 View Post
    for what values of a and b will the parabola y =x^2 + ax + b be tangent to the curve y = x^3 at point (1,1)?

    So we are going to need to solve some equations....

    I am going to call f(x)=y=x^3

    Starting with y=x^2+ax+b we get

    1=1^2+a(1)+b \implies a=-b

    f'(x)=3x^2 \implies f'(1)=3

    Now y'=2x+a using the slope above and the x-coordinate we can solve for a

    3=2(1)+a \implies a =1 so b=-1

    y=x^2+x-1
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  3. #3
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    First you will want to find the derivative of: f(x) = x^3. Using the power rule, f'(x) = 3x^2 Now plug in the 1 to find the slope of the curve at x = 1. 3(1)^2 = 3. Now you will want to find the derivative of the equation g(x) = x^2 + ax + b and set that equal to the derivative of f(x). So g'(x) = 2x + a and f'(x) = 3x^2.
      3x^2 = 2x + a
    Substituting in the 1:
    <br />
\\3 = 2 + a<br />
    <br />
\\a = 1<br />

    Now that you have the a-value, you will want to solve for b. Set f(1) = g(1) to get the b.

    <br />
\\ x^3 = x^2 + x + b<br />
    <br />
\\ 1 = 1 + 1 + b<br />
    <br />
\\ b = -2<br />

    Finally, the equation of the parabola tangent to the cubic is:
    <br />
g(x) = x^2 + x - 1<br />
    Last edited by JoshHJ; February 23rd 2009 at 04:57 PM.
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