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Thread: Finding Equation of a Parabola from Tangent Slopes

  1. #1
    Senior Member topher0805's Avatar
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    Finding Equation of a Parabola from Tangent Slopes

    Find a parabola of the form given below that has slope $\displaystyle m_1$ at $\displaystyle x_1$, slope $\displaystyle m_2$ at $\displaystyle x_2$, and passes through the point $\displaystyle P$.

    $\displaystyle
    y = ax^2 + bx + c$

    $\displaystyle m_1 = 9, x_1 = 10$

    $\displaystyle m_2 = 4, x_2 = 6$

    $\displaystyle P = (9, 2) $
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  2. #2
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    Hello, topher0805!

    Find a parabola of the form $\displaystyle y \:=\:ax^2+bx+c$ .that has slope $\displaystyle 9$ at $\displaystyle x=10$,
    slope $\displaystyle 4$ at $\displaystyle x=6$, and passes through the point $\displaystyle (9,2)$
    The slope is given by: .$\displaystyle y' \:=\:2ax + b$


    $\displaystyle \begin{array}{cccccc}
    x=10,\:y'=9\!: & 2a(10) + b \:=\:9 & \Rightarrow & 20a + b \:=\:9 & {\color{blue}[1]}\\
    x=6,\:y' = 4\!: & 2a(6) + b \:=\:4 & \Rightarrow & 12a + b \:=\:4 & {\color{blue}[2]} \\
    (9,2)\!: &&& 81a + 9b + c \:=\:2 & {\color{blue}[3]}\end{array}$


    Subtract [2] from [1]: .$\displaystyle 8a\:=\:5\quad\Rightarrow\quad\boxed{ a \:=\:\frac{5}{8}}$

    Substitute into [2]: .$\displaystyle 12\left(\frac{5}{8}\right) + b \:=\:4\quad\Rightarrow\quad\boxed{ b \:=\:-\frac{7}{2}}$

    Substitute into [3]: .$\displaystyle 81\left(\frac{5}{8}\right) + 9\left(-\frac{7}{2}\right) + c \:=\:2\quad\Rightarrow\quad\boxed{ c \:=\:-\frac{137}{8}}$


    Therefore: .$\displaystyle \boxed{y \;=\;\frac{5}{8}x^2 - \frac{7}{2}x - \frac{137}{8}}$

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