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Math Help - Equation of a tangent line

  1. #1
    CMH
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    Equation of a tangent line

    I have the following exponential equation
    (x^3 + 2)(x^2 + X)

    Find the equation of the tangent line when x=1

    This is what I do to get the derivative

    (3x^2)(x^2 + x) + (2x + 1)(x^3 + 2)

    When I factor this out I get an answer that includes x^4 and x^3 and x^2 yet the answer for the equation of the line is 15x-9.

    Where am I going wrong? I am really lost on tangent line equations.
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  2. #2
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    Quote Originally Posted by CMH View Post
    I have the following exponential equation
    (x^3 + 2)(x^2 + X)

    Find the equation of the tangent line when x=1

    This is what I do to get the derivative

    (3x^2)(x^2 + x) + (2x + 1)(x^3 + 2)

    When I factor this out I get an answer that includes x^4 and x^3 and x^2 yet the answer for the equation of the line is 15x-9.

    Where am I going wrong? I am really lost on tangent line equations.
    The derivative, at x= 1, will give you the slope of the tangent.

    calculate dy/dx and put x=1 in it to get the slope of tangent.

    slope, \;m  = \left[\frac{dy}{dx}\right]_{x=1}  = [3(1)^2][(1)^2 + 1] + [2(1) + 1][(1)^3 + 2]=15

    After that calculate the y - coordinate of the point, by putting x=1 in the eqn of curve, you will get (1, 6)

    y= [(1)^3 + 2][(1)^2 + 1]= 6

    Now, find the eqn of tangent using point slope form of line,

    y-y_1=m(x-x_1)

    y - 6 = 15 (x - 1)

    y = 15x - 9
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  3. #3
    CMH
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    Thank you

    Now it makes sense. Thank you so much.
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