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Thread: Finding the vertex at a given point?

  1. #1
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    Finding the vertex at a given point?

    Find b and c so that the parabola has vertex .

    What do I do here?
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    Re: Finding the vertex at a given point?

    Use completing the square:
    Convert $\displaystyle y$ to $\displaystyle y=(x-a)^2+q$
    Where $\displaystyle (a,q)$ are the coordinates of the vertex.
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    Re: Finding the vertex at a given point?

    Quote Originally Posted by habibixox View Post
    Find b and c so that the parabola has vertex .

    What do I do here?
    Hi habibixox,

    At the vertex of the parabola the first derivative of y becomes zero. So calculate, $\displaystyle \frac{dy}{dx}$ and equate it to zero. Since the corresponding x value at the vertex is given you can find b. I hope you can do it now.
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    Re: Finding the vertex at a given point?

    wait so is it
    x^2-18x+91?
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    Re: Finding the vertex at a given point?

    using the second method the derivative is
    -8x + x right?
    set that = 0

    0= -8x + x
    0= X(-8+1)
    so final answer is 0?
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  6. #6
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    Re: Finding the vertex at a given point?

    nvm b = 72 now im working on c ill get it i understand it...
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    Re: Finding the vertex at a given point?

    Quote Originally Posted by habibixox View Post
    using the second method the derivative is
    -8x + x right?
    set that = 0

    0= -8x + x
    0= X(-8+1)
    so final answer is 0?
    No.

    $\displaystyle y=-4x^2+bx+c$

    $\displaystyle \frac{dy}{dx}=-4(2x)+b(1)+0$

    $\displaystyle \frac{dy}{dx}=-8x+b$

    Hope you can continue.
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    MHF Contributor Siron's Avatar
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    Re: Finding the vertex at a given point?

    The method I was thinking about is the following.
    Given is the parabola: $\displaystyle y=-4x^2+bx+c$ and the vertex point with coordinates $\displaystyle (9,10)$
    In general we know we can directly read the coordinates of the vertex of a parabola if we convert $\displaystyle y$ to:
    $\displaystyle y=a(x-h)^2+q$ where $\displaystyle (h,q)$ are the coordinates of the parabola.

    First, divide every term by $\displaystyle -4$ so:
    $\displaystyle \frac{y}{-4}=x^2-\frac{b}{4}x-\frac{c}{4}$
    Use completing the square in the RHS:
    $\displaystyle \frac{y}{-4}=\left(x-\frac{b}{8}\right)^2-\frac{b^2}{64}-\frac{c}{4}$
    $\displaystyle \Leftrightarrow y=-4\left(x-\frac{b}{8}\right)^2+\frac{b^2}{16}+c$

    That means:
    (1)$\displaystyle \frac{b}{8}=9 \Leftrightarrow b=72$ (1)
    (2)$\displaystyle \frac{b^2}{16}+c=10$ (2)
    Substituting (1) in (2) gives:
    $\displaystyle 324+c=10 \Leftrightarrow c=-314$

    Therefore the parabola with vertex $\displaystyle (9,10)$ is:
    $\displaystyle y=-4x^2+72x-314$

    But as you see, the method sudharaka wants to use is easier.
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