# stuck in deriving quadratic equation for a parabola

• Feb 28th 2013, 08:59 AM
abdulrehmanshah
stuck in deriving quadratic equation for a parabola
Attachment 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
• Feb 28th 2013, 09:32 AM
HallsofIvy
Re: stuck in deriving quadratic equation for a parabola
It's the graph that is bad- it is NOT precisely a parabola.
• Feb 28th 2013, 09:54 AM
abdulrehmanshah
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
• Feb 28th 2013, 09:56 AM
abdulrehmanshah
Re: stuck in deriving quadratic equation for a parabola
oh i think u are right thanks man
• Feb 28th 2013, 12:59 PM
MINOANMAN
Re: stuck in deriving quadratic equation for a parabola
Abdulrahman

Any curve with formula y=ax^2+bx is a parabola....
• Feb 28th 2013, 04:43 PM
Soroban
Re: stuck in deriving quadratic equation for a parabola
Hello, abdulrehmanshah!

One problem is using rounded decimals.
Use fractions.

Quote:

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

• Feb 28th 2013, 07:04 PM
Prove It
Re: stuck in deriving quadratic equation for a parabola
Quote:

Originally Posted by abdulrehmanshah
Attachment 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.
• Feb 28th 2013, 08:37 PM
abdulrehmanshah
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?????
• Feb 28th 2013, 08:41 PM
abdulrehmanshah
Re: stuck in deriving quadratic equation for a parabola
Quote:

Originally Posted by MINOANMAN
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
• Feb 28th 2013, 08:42 PM
Prove It
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*}
• Feb 28th 2013, 08:43 PM
Prove It
Re: stuck in deriving quadratic equation for a parabola
Quote:

Originally Posted by abdulrehmanshah
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.
• Feb 28th 2013, 09:09 PM
abdulrehmanshah
Re: stuck in deriving quadratic equation for a parabola
Quote:

Originally Posted by Prove It
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
• Feb 28th 2013, 09:46 PM
Paze
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.
• Feb 28th 2013, 11:52 PM
abdulrehmanshah
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
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.