Question 969595
Yes, it is, IF the question says that the range is between minus infinity and plus infinity.  The solutions are exact integers.  

You would need a cubic equation for this, because a quadratic could have the intercepts right, but f(x) as x became very large or very small would be positive or negative infinity.  


{{{graph(300,200,-10,10,-10,10,-(5/4)x^3-(15/4)x^2 +5)}}}

You have another way to do this, so let me explain it better than I did.

It's a cubic, and two intercepts have to be the same, since there are 3 roots and only 2 distinct intercepts.

There can be two roots at 1 and one root at minus 2.  Root at -2 is (x+2), for when (x+2)=0, x=-2.
The root at +1 is (x-1).  We will square that one (use it twice).

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

What is a?  Well, use the y-intercept OR any point they give you.

(0,5) is the point you were given.  Put that into the function.
5=a (-1^2)*2    When x becomes 0 on the right, it is (-1^2)*2) or 2.  
5=2a
a=5/2

You now have f(x)= (5/2)(x-1)^2(x+2)
f(x)=(5/2) * (x^2-2x+1) (x+2)  If I multiply this out

f(x)=(5/2)(x^3-3x+2)

Here is the graph below.

{{{graph(300,200,-10,10,-10,10,(5/2)(x^3-3x+2))}}}

I hope that helps.  I should have done it this way to show you that both functions worked.  You got the 5/4, because you squared the -2 and multiplied it by 1.  I got 5/2 because I squared the -1 and multiplied it by 2.