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Math Help - stuck in deriving quadratic equation for a parabola

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    stuck in deriving quadratic equation for a parabola

    stuck in deriving quadratic equation for a parabola-graphhh.gifi know y intercept is 0 so c =0
    0.36a+0.6b=0
    b=-0.6a
    (0.3^2)a +0.3(-0.6a)=8
    a=-88.9
    b=53.33
    -88.9x^2+53.33x i get x intercepts vertex and area under the curve right but when i plug in a value of x other than intercept or vertex i get a wrong answer need help guys
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    Re: stuck in deriving quadratic equation for a parabola

    It's the graph that is bad- it is NOT precisely a parabola.
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    Re: stuck in deriving quadratic equation for a parabola

    how can u say that do u have like any thing to prove it i mean ur claim that this graph is not a parabola
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    Re: stuck in deriving quadratic equation for a parabola

    oh i think u are right thanks man
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    Re: stuck in deriving quadratic equation for a parabola

    Abdulrahman

    Any curve with formula y=ax^2+bx is a parabola....
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    Re: stuck in deriving quadratic equation for a parabola

    Hello, abdulrehmanshah!

    One problem is using rounded decimals.
    Use fractions.


    Click image for larger version. 

Name:	graphhh.GIF 
Views:	8 
Size:	34.1 KB 
ID:	27296
    I know y-intercept is 0 so c =0.

    The general form is: . y \:=\:ax^2 + bx + c

    You found that c=0\!:\;\;y \:=\:ax^2 + bx

    We know two more points: (0.6,0),\,(0.3,8)

    \begin{array}{cccccccccc}x=0.6,y=0\!:& a(0.6)^2 + b(0.6) \:=\:0 & \Rightarrow & 0.36a + 0.6b \:=\:0 & [1] \\ x=0.3,y=8\!: & a(0.3)^2 + b(0.3) \:=\:8 & \Rightarrow & 0.09a + 0.3b \:=\:8 & [2] \end{array}

    \begin{array}{cccccc}\text{Multiply [2] by -2:} & \text{-}0.18a - 0.6b &=& \text{-}16 \\ \text{Add [1]:} & 0.36a + 0.6b &=& 0 \end{array}

    We have: . 0.18a \:=\:\text{-}16 \quad\Rightarrow\quad a \:=\:\text{-}\frac{16}{0.18} \:=\:\text{-}\frac{1600}{18} \quad\Rightarrow\quad a \:=\:\text{-}\frac{800}{9}

    Substitute into [1]: . 0.36\left(\text{-}\tfrac{800}{9}\right) + 0.6b \:=\:0 \quad\Rightarrow\quad -32 + 0.6b \:=\:0

    . . . . . . . . . . . . . . . 0.6b \:=\:32 \quad\Rightarrow\quad b \:=\:\frac{32}{0.6} \:=\:\frac{320}{6} \quad\Rightarrow\quad b \:=\:\frac{160}{3}


    Therefore: . y \;=\;\text{-}\frac{800}{9}x^2 + \frac{160}{3}x

    Thanks from Paze
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    Re: stuck in deriving quadratic equation for a parabola

    Quote Originally Posted by abdulrehmanshah View Post
    Click image for larger version. 

Name:	graphhh.GIF 
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ID:	27296i know y intercept is 0 so c =0
    0.36a+0.6b=0
    b=-0.6a
    (0.3^2)a +0.3(-0.6a)=8
    a=-88.9
    b=53.33
    -88.9x^2+53.33x i get x intercepts vertex and area under the curve right but when i plug in a value of x other than intercept or vertex i get a wrong answer need help guys
    Since we can easily read off the turning point at (0.3, 8), I would go straight for substituting into the turning point form of the quadratic.

    \displaystyle \begin{align*} y &= a \left( x - h \right) ^2 + k \\ y &= a \left( x - 0.3 \right) ^2 + 8 \end{align*}

    You can read off several other points which lie on the curve. Substitute one in and solve for the value of a. Then you will have the equation of your quadratic.
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    Re: stuck in deriving quadratic equation for a parabola

    i got a =-100 and when i plug in x as 0 i dont f(x)=0?????
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    Re: stuck in deriving quadratic equation for a parabola

    Quote Originally Posted by MINOANMAN View Post
    Abdulrahman

    Any curve with formula y=ax^2+bx is a parabola....
    actually this is a graph the examiner asks me to find the area under the curve it never said or hinted that is a parabola
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    Re: stuck in deriving quadratic equation for a parabola

    You can use the point (0, 0) if you like, but that won't give you a = -100.

    \displaystyle \begin{align*} y &= a \left( x - 0.3 \right)^2 + 8 \\ 0 &= a \left( 0 - 0.3 \right)^2 + 8 \\ 0 &= a \left( -0.3 \right)^2 + 8 \\ 0 &= 0.09\, a + 8 \\ -8 &= 0.09\, a \\ -\frac{8}{0.09} &= a \\ -\frac{800}{9} &= a  \end{align*}
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    Re: stuck in deriving quadratic equation for a parabola

    Quote Originally Posted by abdulrehmanshah View Post
    actually this is a graph the examiner asks me to find the area under the curve it never said or hinted that is a parabola
    To find the area under the curve, you will need to evaluate a rule for it so that you can integrate it.
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    Re: stuck in deriving quadratic equation for a parabola

    Quote Originally Posted by Prove It View Post
    You can use the point (0, 0) if you like, but that won't give you a = -100.

    \displaystyle \begin{align*} y &= a \left( x - 0.3 \right)^2 + 8 \\ 0 &= a \left( 0 - 0.3 \right)^2 + 8 \\ 0 &= a \left( -0.3 \right)^2 + 8 \\ 0 &= 0.09\, a + 8 \\ -8 &= 0.09\, a \\ -\frac{8}{0.09} &= a \\ -\frac{800}{9} &= a  \end{align*}
    take point 0.1
    4=a(0.1-0.3)^2+8
    now we get a=-100??? why
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    Re: stuck in deriving quadratic equation for a parabola

    Abdulrehmanshah,

    In order to find the area under your graph, you need to be familiar with integral calculus. Are you sure that you are trying to find the area under the curve? An example of that can be found here: http://mathin3d.com/Calculus_Integral_v01_01_00004.jpg
    It's just that your title states that you need to derive an equation but the thread seems to revolve around reading the equation from the graph..And now you need to find an area?

    Just in case of a potential communication problem.
    Last edited by Paze; February 28th 2013 at 08:52 PM.
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    Re: stuck in deriving quadratic equation for a parabola

    yea thats because i thought that it is a quadratic equation cause every parabola has a quadratic equation but now i realize i was wrong so what equation should be the equation for this curve so that i could find the area and yes i do know how to find the area by integrating but what would i integrate if i dont have the equation>???
    Quote Originally Posted by Paze View Post
    Abdulrehmanshah,

    In order to find the area under your graph, you need to be familiar with integral calculus. Are you sure that you are trying to find the area under the curve? An example of that can be found here: http://mathin3d.com/Calculus_Integral_v01_01_00004.jpg
    It's just that your title states that you need to derive an equation but the thread seems to revolve around reading the equation from the graph..And now you need to find an area?

    Just in case of a potential communication problem.
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