How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (x+5)/(x+3)#?

1 Answer
Jan 1, 2017

Answer:

Horizontal: #larr y - 1 rarr#.
Vertical : #uarr x = -3 darr#.
See Socratic graph, with asymptotes.

Explanation:

By actual division,

#y = f(x)=Q+R/S=1+2/(x+3)#

y = Q gives parabolic, ( slant-straight) oblique or horizontal

asymptote according as Q is quadratic, linear or constant,

respectively.

x = zero(s) of S give vertical asymptote(s).

Here,

they are #y = 1 and x = -3#.

In the graph, the asymptote #x=-3# has gone into hiding. I am

unable to get it out. However, y = 1 is marked.

The graph is the a rectangular hyperbola (y-1)(x+3)=2, with center at

the meet of the asymptotes, #(-3, 1)#.

graph{(y-1)(x+3)(y-(x+5)/(x+3))=0 [-11.71, 11.71, -5.85, 5.86]}