f(x) = (3x+3)/(2x+4) = (3(x+1))/(2(x+2)) = 3/2*(x+1)/(x+2)f(x)=3x+32x+4=3(x+1)2(x+2)=32⋅x+1x+2
=3/2*(x+2-1)/(x+2)=32⋅x+2−1x+2
=3/2((x+2)/(x+2)-1/(x+2))=32(x+2x+2−1x+2)
=3/2(1-1/(x+2))=32(1−1x+2)
=3/2-3/(2(x+2))=32−32(x+2)
with exclusion x != -2x≠−2
So as x->-oox→−∞ the term 3/(2(x+2)) ->0_-32(x+2)→0− so f(x)->(3/2)_+f(x)→(32)+
As x->(-2)_-x→(−2)− the term 3/(2(x+2))->-oo32(x+2)→−∞ so f(x)->oof(x)→∞
As x->(-2)_+x→(−2)+ the term 3/(2(x+2))->oo32(x+2)→∞ so f(x)->-oof(x)→−∞
As x->oox→∞ the term 3/(2(x+2)) -> 0_+32(x+2)→0+ so f(x)->(3/2)_-f(x)→(32)−
The intersection with the xx axis occurs at (-1, 0)(−1,0) since the numerator of the original expression is 00 for x = -1x=−1
The intersection with the yy axis can be found by substituting x=0x=0 into the original equation to derive (0, 3/4)(0,34)
graph{ (3x+3)/(2x+4) [-12.29, 7.71, -3.44, 6.56]}
