Question 706591
(a) direction of opening:
Depends on the sign of a.  If a<0 Then opens downward; if a>0 then opens upward.


(b) location of vertex:
Best to complete the square to put into standard form allowing vertex point to be read directly from the equation.  Since we can see or read the roots from the factored equation given, we know that the axis containing the vertex will be in the middle of x=r and x=s.

y=a(x^2-(r+s)x+rs)
Inside the outer parentheses, you want to add {{{(r+s)^2/2^2}}}, and then outside those parentheses you want to subtract {{{a((r+s)^2)/2^2}}}.
.
.
On paper, I find vertex, ({{{(r+s)/2}}},{{{(ars-a(r+s)^2/(4))}}})


(e) Y intercept:
Try let x=0.  y=a(0-r)(0-s)={{{+ars}}}
The y intercept is (0,ars)