This question is designed to test shifting and scaling. It is a great question that applies to physics and real life.
You can start off with y=x^2. This gives you a generic U shaped graph.
To flip graph, make X negative...y=x^2
To make this graph display the max height of 9, you would shift the graph upward by subtracting 9 from y... y9=(x^2)...y=(x^2)+9
Now to get a distance of 20 you would stretch the graph, you do this by multiplying or dividing x by a number.
To find the number(we can use "c") I put the desired points into the equation.. 0=(c*10^2)+9...c=.09
Now we plug c into the equation...y=(.09x^2)+9
This equation displays the flight of the kick.
The last step is to set the graph to start at the origin.To do this we must shift the graph to the right. We do this by subtracting 10 from x...
y=[.09(x10)^2]+9
This may seem like a long process but I tried to go slow. Basically you shift and scale(strech/squeeze) you graph by manipulating you equation just as I did.
Shift up/down...subtract/add to y
Shift right/left...subtract/add to x
Stretch/squeeze... multiply factor to x ... this is the hardest part for me. I usually find the factor to multiply by plugging in desired points.
11/14/2012

Alex O.
Comments
It's not at all clear how you are solving for your 'c' variable... You start with an inverted parabola y=x^{2} + 9, and somehow you end up with "c*10^2+9=0"?? I'm not sure what you are doing there. What is "c*"? Where did 10^2 come from?
Also, your final equation doesn't satisfy the requirement of a maximum height of 9, but I think it's because you just typed the wrong number... I think your final equation (the one that is bolded) was meant to be:
y = [.09(x10)^{2}]+9
I like the original ideas. It is clear to me.
The ideas is to use transformation to get the answer.
You start with y = ax^2+9, which means the axis of symmetry is yaxis.
When x = 10, y = 0, thus you get a = 9/100.
So, y = (9/100)x^2 + 9 <==This is a correct answer.
If you want to start with x = 0, then you need to shift parabola to the right by 10.
Therefore, the answer becomes y = (9/100)(x10)^2 + 9, which is equivalent to my answer y = (9/100)(x)(x20).
Yes you are right. I meant to put + 9. Your bolded equation is what I meant.
The part with getting "c" was unclear so I will try again.
We have to strech the graph to get the desired distance of the kick. To strech a graph to have to multiply x by some factor. I just chose "c" to represent that factor. You could have picked any letter to represent the factor.
To find it I plugged in the desired values of y=0 and x=10 into the equation. That is where the 10^{2} came from. I think that is where I lost you. I'll try to explain it better.
I chose these values because my max height, at that time, was at x=0. We know that the max height will be reached when half the total distance is travelled. So I need 10 units(half the distance) before and after the max height. Y=0 represents the kick hitting the ground. So after the max height, y(the height) should be zero at x=10(this would represent the second half of total travel). This would give ten units on each side of the max height and 20 units in between the two instances when y=0(start to finish).
So now that I have royaly confused you again, we can find c by plugging in y=0 at x=10.
y=(c)*x^{2}+9 ... again c is just that factor we are trying to find. Now plug in 0 for y and 10 for x and we get... 0=c*10^{2}+9 ... solve for c and get... c=9/100=0.09
Now we plug c back into the original equation.. y=0.09*x^{2}+9.. This equation has a max height of 9 and total distance of 20. But the max height is still at the x=0. From here we just shift the graph to the right to get the equation to start at the origin.
I hope this helps!
Sorry Robert it is the same answer. I commented before you changed the typo.
I actually didn't remember how I was taught to do this so I just went to the basics of shifting and scaling.
Alex, I think it's just your notation that is confusing... the * in your equations is very unconventional. But I think I follow now.. I think that when you write:
c*x^{2}+9 (and similar in multiple places)
What you really mean is
(c)(1)(x^{2})+9
or better: c*x^{2}+9