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Math Help - Parabola Question! Help!

  1. #1
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    Thumbs down Parabola Question! Help!

    The parabola has the turning point at (z, -8) and y-intercept at 10, and an x-intercept at x=5.

    What is --- value of z?

    The equation of parabola?

    Plz help!
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  2. #2
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    Quote Originally Posted by mibamars View Post
    The parabola has the turning point at (z, -8) and y-intercept at 10, and an x-intercept at x=5.

    What is --- value of z?

    The equation of parabola?

    Plz help!
    There are several approaches that can be taken. This is one of them:

    The equation can be written in turning point form as y = a(x - z)^2 - 8.

    Substitute the known points (0, 10) and (5, 0) to get two equations in the unknowns a and z. Solve these two equations simultaneously to get the values of a and z.
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  3. #3
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    Still cant solve it. Stuck on the unknown values. Can someone do a full solve?
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    There are several approaches that can be taken. This is one of them:

    The equation can be written in turning point form as y = a(x - z)^2 - 8.

    Substitute the known points (0, 10) and (5, 0) to get two equations in the unknowns a and z. Solve these two equations simultaneously to get the values of a and z.
    Quote Originally Posted by mibamars View Post
    Still cant solve it. Stuck on the unknown values. Can someone do a full solve?
    y = a(x - z)^2 - 8

    Plug in the two points:
    10 = a(-z)^2 - 8
    and
    0 = a(5 - z)^2 - 8
    respectively.

    The first equation says:
    10 = az^2 - 8
    so
    a = \frac{18}{z^2}

    Inserting this into the second equation gives:
    0 = \left ( \frac{18}{z^2} \right )(5 - z)^2 - 8

    0 = (5 - z)^2 - 8

    Solve this for z, then use that z value to find a.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    y = a(x - z)^2 - 8

    Plug in the two points:
    10 = a(-z)^2 - 8
    and
    0 = a(5 - z)^2 - 8
    respectively.

    The first equation says:
    10 = az^2 - 8
    so
    a = \frac{18}{z^2}

    Inserting this into the second equation gives:
    0 = \left ( \frac{18}{z^2} \right )(5 - z)^2 - 8

    0 = (5 - z)^2 - 8 Mr F says: There's a little typo here. Rather than fix it, I'm gonna suggest you (mibamars, NOT topsquark ) pick things up from the line above. There's a bit of algebra to go but all the thinking part of the question has been done ......

    Solve this for z, then use that z value to find a.

    -Dan
    ..
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  6. #6
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    Hello, mibamars!

    Another approach . . .


    The parabola has the turning point at (z,\,\text{-}8),
    and y-intercept at 10, and an x-intercept at 5.

    Find z and the equation of the parabola.

    The general equation of a parabola is: . y \;=\;ax^2 + bx + c


    We are given three points on the parabola.

    (0,10)\!:\;\;10 \:=\:a(0^2) + b(0) + c\quad\Rightarrow\quad\boxed{ c \:=\:10}

    (5,0)\!:\;\;0 \:=\:25a + 5b + 10\quad\Rightarrow\quad 5a + b \:=\:\text{-}2\;\;{\color{blue}[1]}

    (z,\text{-}8)\!:\;\;\text{-}8 \:=\:az^2 + bz + 10\quad\Rightarrow\quad az^2 + bz \:=\:\text{-}18\;\;{\color{blue}[2]}


    We are told that: . y'(z) = 0
    We have: . y' \:=\:2ax + b\quad\hdots\quad\text{Hence: }\; 2az + b\:=\:0\;\;{\color{blue}[3]}


    \begin{array}{cccc}\text{Multiply {\color{blue}[3]} by }z\!:& 2az^2 + bz &=&0 \\<br />
\text{Subtract {\color{blue}[2]}}\!: & az^2 + bz &=&\text{-}18 \end{array}
    And we have: . az^2 \:=\:18\quad\Rightarrow\quad a \:=\:\frac{18}{z^2}\;\;{\color{blue}[4]}

    Substitute into [3]: . 2\left(\frac{18}{z^2}\right)z +b \:=\:0\quad\Rightarrow\quad b \:=\:-\frac{36}{z}\;\;{\color{blue}[5]}

    Substitute [4] and [5] into [1]: . 5\left(\frac{18}{z^2}\right) - \frac{36}{z}\:=\:-2\quad\Rightarrow\quad z^2 - 18z + 45\:=\:0

    . . and we have: . (z-3)(z-15)\:=\:0\quad\Rightarrow\quad\boxed{z \:=\:3,\:15}


    There are two solutions . . .

    . . \begin{array}{ccccccccc}z \,=\,3, & a \,=\,2, & b\,=\,\text{-}12 && \Longrightarrow & &\boxed{ y \;=\;2x^2 - 12x + 10} \\ \\<br /> <br />
z\,=\,15 & a \,=\,\frac{2}{25}, & b\,=\,\text{-}\frac{12}{5} && \Longrightarrow &&\boxed{ y \;=\;\frac{2}{25}x^2 - \frac{12}{5}x + 10} \end{array}

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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Mr F says: There's a little typo here.
    Oopsies! Thanks for the catch. (That darned distributive law!)

    -Dan
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