Constructing a quadratic equation

Question

So basically I have been assigned a question that involves constructing a quadratic equation from scratch and graphing it. So here are the details.

We are designing a water arc fountain, and it has a maximum of $20$ feet wide and taller than $6$ feet and shorter than $50$ feet. We are first asked to place it in a first quadrant graph and have the surface of the pool correspond to the x-axis with the left side at the origin. Using the equation $y = a(x-x_1)(x-x_2)$, where $(x_1, 0)$, $(x_2, 0)$ are the $x$ intercepts of the parabola, we have to choose a value of $a$ that will produce a reasonable arc. Then we have to convert the equation to the form $y = ax^2+bx+c$. Also, how high will the water arc be if we move the beginning point $1$ foot to the right.

I know that this question might be a piece of cake to do, but I just couldn't wrap my head around it. Would greatly appreciate an answer ASAP.

Does this mean $x_1=0$ and $x_2=20$? You say the pool is $20$ feet wide but it is unclear whether the water is supposed to arc all the way from one edge to the other or somewhere inside the pool. — David K, Feb 28 '15 at 17:01
I'm sorry, it has a maximum of 20 feet, not 20 feet. This was a mistake from my side. — Majid Alshamsi, Feb 28 '15 at 17:03
So $x_1=1$, $x_2=19$ would be OK? Or even $x_1=5$, $x_2=15$? — David K, Feb 28 '15 at 17:05
The water arc should start from origin. God what is wrong with me. — Majid Alshamsi, Feb 28 '15 at 17:05

score 0 · Accepted Answer · answered Feb 28 '15 at 17:15

HINT: The highest point on the parabola is halfway between $x_1$ and $x_2$ (the two $x$-intercepts). This is a property of any parabola of the form $y = ax^2+bx+c$ (provided that the parabola has two $x$-intercepts, of course).

Because of this fact, once you have chosen a suitable $x_1$ and $x_2$, you can select $x$ halfway between $x_1$ and $x_2$ and evaluate $y = a(x-x_1)(x-x_2)$ there. The result will still have a factor of $a$ in it because you haven't yet decided what $a$ should be, but now it should not be so hard to find a suitable $a$ so that when you substitute that value, the maximum $y$ is between $6$ and $50$ as desired.

So for example if I chose x1=0 and x2=10, would -1 be a reasonable value for a? — Majid Alshamsi, Feb 28 '15 at 17:20

Constructing a quadratic equation

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