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Math Help - Finding the values of a, b and c.

  1. #1
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    Finding the values of a, b and c.

    The tangent to the curve of y=ax^2 + bx +c at the point where x=2 is parallel to the line y=4x. There is a stationary point at (1, -3). Find the value of a, b and c.

    I was really unsure of how to work out this question, so first I differentiated y.
    dy/dx=2ax+b
    The point (1, -3) is on the curve and therefore:
    -3=a(1)^2+b(1)+c
    -3=a^2+b+c

    But I don't know how to find another equation, and I honestly don't even know if what I've done so far is on the right track.
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  2. #2
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    Re: Finding the values of a, b and c.

    Hello, grooverandshaker!

    You were off to a good start . . .


    \text{The tangent to the curve of }y\:=\:ax^2 + bx +c\text{ at the point where }x=2
    . . \text{is parallel to the line }y\,=\,4x.
    \text{There is a stationary point at }(1,-3).
    \text{Find the values of }a, b\text{ and }c.

    \text{The slope of the line }y\,=\,4x\text{ is }4.

    The slope of a tangent to the curve is: . y' \:=\:2ax + b
    When x = 2, the slope of the tangent is 4.
    . . \text{We have: }\:4a + b \:=\:4

    There is a stationary point at (1,-3).
    That is, when x = 1, the derivative equals zero.
    . . We have: . 2a + b \:=\:0

    The point (1,-3) is on the curve.
    That is, when x = 1,\:y = -3
    . . We have: . a + b + c \:=\:-3


    We have a system of equations: . \begin{Bmatrix}4a + b\qquad &=& 4 & [1] \\ 2a + b \qquad &=& 0 & [2] \\ a + b + c &=& \text{-}3 & [3] \end{Bmatrix}

    Subtract [1] - [2]: . 2a \,=\,4 \quad\Rightarrow\quad \boxed{a \,=\,2}

    Substitute into [2]: . 4 + b \:=\:0 \quad\Rightarrow\quad \boxed{b \:=\:-4}

    Substitute into [3]: . 2 - 4 + c \:=\:-3 \quad\Rightarrow\quad \boxed{c \:=\:-1}


    Therefore, the function is: . y \:=\:2x^2 - 4x - 1

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  3. #3
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    Re: Finding the values of a, b and c.

    Thank you so much mate! That makes perfect sense. =)
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