Writing equations for parabolas?

The information tells you that the vertex of the parabola is the highest point of the graph, at (20, 15) on the coordinate plane. Graph that point. The x-values and y-values represent the measurement units: yards. The parabola is chopped off once you go down to the x-axis on the left side of the vertex or the right side of it.

Graph the point (0,0) and realize that the vertex is halfway between the two points that are on the x-axis, so graph the point ( (twice of 20), 0) which is (40,0). Connect the two points on the x-axis with the vertex by drawing a curve between (0,0) and (20,15) and a curve between (20,15) and (40,0).

The y and x axes are to be labeled with yards as the units.

HOW TO GET THE EQUATION: If the parabola shifted its vertex to (0,0), the two other points would be shifted to (-20,-15) and (20, -15). If x-value in these coordinates gets squared and multiplied by a and added to b times x in order to get the y-value, you would get 400a - 20b = -15 and 400a + 20b = -15. Since you need an a-value for a parabola's equation, and the only way those two equations can be true is if b=0, set b=0 and solve for a's value with the equation 400a = -15. a = -3/80.

So if the parabola was shifted like I described, the equation would be y = -3/80 * x^2. But to undo the shift, change it to y = -3/80 * (x-20)^2 + 15 and write it next like -3/80 * (x^2 - 40x + 400) + 15.

Equation is (-3/80)x^2 + (3/2)x.

So the vertex is given for the maximum point on the parabola. If you graph it from its Y point as the height (15 yards) and horizontal distance (20 yards) for x. The equation should look like:

y = (x-20) + 15

Please note that John's equation is wrong. There is no possible way that the equation can be a parabola.