Im having a brain fart.
I know that the formula is
$\displaystyle y=ax^2+bx+c$
But what do the letters represent again. Also, how is slope calculated on a parabola?
Ummm... I think you have forgotten more than that to which you are admitting.
x and y are variables related to a Cartesian Coordinate System
a, b, and c are parameters that require additional information in order to pin down. They can be virtually anything. They do not have specifc names or purposes unless you specify a particular application or purpose.
"slope", as a single, fixed entity, has no meaning for a parabola. It changes all the time.
So, for real, what exactly was your question? Perhaps a problem statement would be beneficial?
Wow.. everyone here makes everything too complicated but ok, ill simplify it. Heres the question. Its a graph and i have to find the equation, heres the table of values in ordered pairs
x= Time (s)
Y= velocity (l/s)
(0,0)
(.25,0.22)
(.5,0.45)
(.75,0.61)
(1,0.75)
(1.25,0.83)
(1.5,0.85)
(1.75,0.83)
(2,0.75)
(2.25,0.61)
(2.5,0.45)
2.75,0.22)
(3.0,0)
After i find the equation i also have to find how many seconds have paassed when her velocity is .5
No, you simply did not provide sufficient information. You did not say anything about a model of time and velocity as a parabola. You just said "parabola".
Now that you have provided useful information, this is an easily solved problem. Please make a practice of providing sufficient information up front, then you will not see things that are "too complicated". It is your previous attempt at simplifying that is the communication problem. Just provide the problem and show your work. That's all there is to it.
In the standard model, we have $\displaystyle y(t) = -(1/2)gt^{2} + v_{0}t + h_{0}$
Where:
y = the height of whatever object we are watching
t = time (usually in seconds)
g = Acceleration due to Gravity (Taken as positive to let the negative sign out front emphasize the direction.)
$\displaystyle v_{0}$ = Initial Velocity
$\displaystyle h_{0}$ = Initial Height
This brings us to the "Show Your Work" part.
If we can assume that all of your data lie exactly on the parabola you need to start with the equation $\displaystyle y = ax^2 + bx + c$. You need three points to determine the values of a, b, and c.
So pick any three points in your data set, say the first three. Then
$\displaystyle 0 = a \cdot 0^2 + b \cdot 0 + c$
$\displaystyle 0.22 = a \cdot 0.25^2 + b \cdot 0.25 + c$
$\displaystyle 0.45 = a \cdot 0.5^2 + b \cdot 0.5 + c$
Solve this system of equations and you have your a, b, and c.
If the data cannot be assumed to lie on the parabola, ie. this is a data set from a real experiment, then you need to do a quadratic regression on your data. You can do this on most advanced TI calculators (the 89 and 92 at least, not to mention the Voyager), or I know it can be done in Excel as well.
-Dan
Yikes. It's good old reality day, isn't it?
Yes, it does work. Your evaluation is not a very good one. Please try to explain what it is that you need. The tricky part is knowing enough to ask the question. When you demonstrate a complete lack of necessary backgroud, it will be very frustrating for everyone - as can be seen.
I get y = -0.0239t^2 + 0.3352t - 0.3327. It's not quite a parabola. It looks to me like the first ones from the borders are a little out of line - among other slight adjustments.