determine constant a and function f if f(1)=5 and f(x+y)-f(x)=axy-2y^2?

The notation, f(x), pronounced "f of x", represents the
output of a function, f, when the value of x is inputed.  For instance, f(x) = x^2-4 has
a value of 0 when x=2, or f(2) = 2^2-4 = 0.

The function in
this question has two facts known about it.  f(1)=5, which means the value of f(x) = 5
when x=1. The other fact is:

f(x+y) -f(x) =
axy-2y^2

Substituting a 1 for
x:

f(1+y) - f(1) =
a(1)y-2y^2

Substitute f(1) = 5: f(1+y) - 5 =
ay-2y^2

Add 5 to both sides: f(1+y) =
ay-2y^2+5

Rearrange the terms: f(y+1) =
-2y^2+ay+5

Factor out a -2 on the right:  f(y+1) =
-2(y^2-(a/2)y+____)+5

Knowing the blank above needs to be a
1 to get (y+1)^2:

f(y+1) = -2(y+1)^2+5+2 :Compensate for
the -2*1

f(y+1) = -2(y+1)^2 +
7

Therefore, f(x)=-2x^2+7 and
a=-4.

Since (y+1)^2 = y^2 + 2y +1, -a/2 = 2 and
a=-4.

Proof: f(1) = -2(1)^2 + 7 =5