Graphing

First, let's write this in vertex form. 

We have $y = a(x-h)^2 + k$ where $(h,k)$ denotes the vertex. From the graph, we can write, $y=a(x-3)^2+1$ since the vertex is at (3, 1)

Now, we have to find a. We can use the y intercept to help. We have 

$7 = a (0 - 3)^2 + 1 \\ 6 = 9a \\ a = 6/9 = 2/3$

We find that when $y=2/3(x-3)^2+1$, we have a y intercept at (0, 7), meaning we found our equation. 

Now, we simply have to expand it. Expaning, we get $y = \frac{2}{3} x^ 2 − 4 x + 7$. 

This means $abc = 2/3(-4)7 = -56/3$

-56/3 is our answer!

Thanks! :)

We have the vertex form

y = a(x - h)^2  + k         where the vertex  (h,k)  = (3,1)

We  know that  (0,7)  is on the graph..so....

7  = a (0 - 3)^2  + 1

6 =  9a

a = 6/9  = 2/3

So

y = (2/3) ( x - 3)^2 + 1

y = (2/3) (x^2 - 6x + 9) + 1

y = (2/3)x^2 - 4x +6 + 1

y = (2/3)x^2 - 4x + 7

a*b*c =  (2/3) (-4) (7)  =  -56 / 3