Equation for a parabola

[norton]

I have a graph of a parabola that opens downwards with the Vertex (5,2) and crossing at points (3,0) and (7,0).
I used the forumla y=ax^2 + bx + c
So I worked it out like this:
y= a (x-3)(x-7)
2= a (5-3)(5-7)
2= a (2)(-2)
2= a(-4)
a= -1/2
y= -1/2(x-3)(x-7)
y= -1/2(x^2 - 10x + 21)
y= -1/2x^2 -5x - 10.5

How ever my choices for answers are

2y= 4 - (x-5)^2
y= 2 -(x-5)^2
2y= 10 - (x-2)^2
y= 5 - (x-2)^2[/img]

 

[galactus]

Your answer is OK, good work. Except, you have a negative where a positive should be.

\(\displaystyle \L\\y=\frac{-x^{2}}{2}+5x-\frac{21}{2}\)

Simplified, this is equivalent to the second choice.

Complete the square on your answer and you should get it.