Question 404617
The horizontal asymptotes of {{{y = 4/(x-2) + 3}}} occur when {{{lim(x->infinity, 4/(x-2) + 3)}}} or {{{lim(x->-infinity, 4/(x-2) + 4)}}} are defined (this is just the formal way of saying that when x gets really large, the value of y is a finite number).


In this case, since the degree in the denominator is larger, {{{lim(x->infinity, 4/(x-2)) = 0}}} and {{{lim(x->-infinity, 4/(x-2)) = 0}}}. Since the expression converges, we can add three to each limit and obtain 3 as our horizontal asymptotes. We can check by graphing the function, if we wish:


{{{graph(300, 300, -20, 20, -20, 20, 4/(x-2) + 3, 3)}}}