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Math Help - Tangent of a point

  1. #1
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    Tangent of a point

    I attached the question number. I found the answer to A and B but I'm stuck on C.

    For B I got y=x+7

    but the tangent line for equation is 3x^2 -2x -4

    Thank you.
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  2. #2
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    Re: Tangent of a point

    Oh, dear. 3x^2- 2x- 4 is a parabola, NOT any line at all! What you mean is that the derivative of x^3- x^2- 4x+ 4 so gives the slope of the tangent line.

    Given that, at x= a, the derivative is 3a^2- 2a- 4 and the value is a^3- a^2- 4a+ 4. That means that the equation of the tangent line, at x= a, is y= (3a^2- 2a- 4)(x- a)+ a^3- a^2- 4a+ 4. Saying that line goes through (0, -8) mean that x= 0, y= -8 makes that a true equation: solve (3a^2- 2a- 4)(0- a)+ a^3- a^2- 4a+ 4= -2a^3+ a^2+ 4= -8.
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  3. #3
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    Re: Tangent of a point

    Quote Originally Posted by minneola24 View Post
    I attached the question number. I found the answer to A and B but I'm stuck on C.

    For B I got y=x+7

    but the tangent line for equation is 3x^2 -2x -4

    Thank you.
    for c,
    remember that (a,b) and (0,-8) lie on the same line,i.e the tangent line. So the slope of the line is  \frac{8+b}{a}.
    Again you know that the slope of the 3x^2-2x-4(which you got by differentiating ) and thus for (a,b) the slope is 3a^2-2a-4.
    equate the slopes:
    \frac{8+b}{a}=3a^2-2a-4 \implies 8+b=3a^3-2a^2-4 and b=a^3-a^2-4a-4 from the original equation:
    so 8+a^3-a^2-4a-4=3a^3-2a^2-4a  \implies 2a^3-a^2-12=0 hence a=2 and b=2^3-2^2-4(2)+4=0
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