Question 590894
Let's rewrite the function f(x)=(2x^2+3x+1)/(2x+1), since we can factor 2x^2+3x+1 into (2x+1)(x+1). We can always cancel the factor of 2x+1, except at -1/2 where the function is undefined (since we can't divide by zero). Thus, f(1)=1+1=2, f(0)=0+1=1, and f(-1)=-1+1=0.