An asymptote of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity. Horizontal asymptotes are parallel to the x-axis (fixed y number) that the graph of the function approaches as x tends to +∞
or −∞. Vertical asymptotes are vertical lines (perpendicular to the x-axis, fixed x number) near which f(x) tends to +∞ or −∞.
a) Find the vertical asymptote of the function f(x) = 4 over x + 5 = 4/(x+5)
Vertical, we are looking for an x number where f(x) -> ∞. In this case it happens when denominator=0, f(x)->∞, which is when x=-5.
b) Find the horizontal asymptote of the function g(x) = 5x^2 -4 over x + 1=(5x2-4)/(x+1)
Horizontal, we are looking for the g(x) when x tends to ∞. When x increases, g(x) -> 5x ( the -4 and +1 matter much less with x->∞, we ignore them). In this case, g(x) increases with x and does not -> to a fixed number (horizontal line). Therefore, there
is no horizontal asymptote for this function.
c) Find the vertical and horizontal asymptote of the function f(x) = 3x – 1 over x +4=(3x-1)/(x+4)
Vertical, f(x) -> ∞, that happens again when denominator=x+4=0, therefore the line is x=-4.
Horizontal, x -> ∞, f(x) -> 3x/x=3 (again, with x -> ∞, -1 and +4 matter very little). The line is f(x)=3
d) Find the vertical and horizontal asymptote of the function g(x) = _x + 7_ over x2 - 4=(x+7)/(x2-4)
Vertical, g(x) -> ∞, that happens again when denominator=x2-4=0, and x=2 or x=-2, these are two vertical lines.
Horizontal, x -> ∞, g(x) ->x/x2=1/x, which ->0 with x -> ∞, therefore the horizontal line is g(x)=0.
Hope this helps.