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

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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