Hello, abdulrehmanshah!
One problem is using rounded decimals.
Use fractions.
The general form is: .$\displaystyle y \:=\:ax^2 + bx + c$
You found that $\displaystyle c=0\!:\;\;y \:=\:ax^2 + bx$
We know two more points: $\displaystyle (0.6,0),\,(0.3,8)$
$\displaystyle \begin{array}{cccccccccc}x=0.6,y=0\!:& a(0.6)^2 + b(0.6) \:=\:0 & \Rightarrow & 0.36a + 0.6b \:=\:0 & [1] \\ x=0.3,y=8\!: & a(0.3)^2 + b(0.3) \:=\:8 & \Rightarrow & 0.09a + 0.3b \:=\:8 & [2] \end{array}$
$\displaystyle \begin{array}{cccccc}\text{Multiply [2] by -2:} & \text{-}0.18a - 0.6b &=& \text{-}16 \\ \text{Add [1]:} & 0.36a + 0.6b &=& 0 \end{array}$
We have: .$\displaystyle 0.18a \:=\:\text{-}16 \quad\Rightarrow\quad a \:=\:\text{-}\frac{16}{0.18} \:=\:\text{-}\frac{1600}{18} \quad\Rightarrow\quad a \:=\:\text{-}\frac{800}{9}$
Substitute into [1]: .$\displaystyle 0.36\left(\text{-}\tfrac{800}{9}\right) + 0.6b \:=\:0 \quad\Rightarrow\quad -32 + 0.6b \:=\:0 $
. . . . . . . . . . . . . . . $\displaystyle 0.6b \:=\:32 \quad\Rightarrow\quad b \:=\:\frac{32}{0.6} \:=\:\frac{320}{6} \quad\Rightarrow\quad b \:=\:\frac{160}{3}$
Therefore: .$\displaystyle y \;=\;\text{-}\frac{800}{9}x^2 + \frac{160}{3}x$
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.
$\displaystyle \displaystyle \begin{align*} y &= a \left( x - h \right) ^2 + k \\ y &= a \left( x - 0.3 \right) ^2 + 8 \end{align*}$
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.
You can use the point (0, 0) if you like, but that won't give you a = -100.
$\displaystyle \displaystyle \begin{align*} y &= a \left( x - 0.3 \right)^2 + 8 \\ 0 &= a \left( 0 - 0.3 \right)^2 + 8 \\ 0 &= a \left( -0.3 \right)^2 + 8 \\ 0 &= 0.09\, a + 8 \\ -8 &= 0.09\, a \\ -\frac{8}{0.09} &= a \\ -\frac{800}{9} &= a \end{align*}$
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.
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>???