Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = (x+1) / (x+2)`?

Answer 1

Vertical : `uarr x=-2 darr`.Horizontal : `larr y=1 rarr`. See this asymptotes-inclusive rectangular hyperbola in the Socratic graph.

Explanation:

`y = f(x)=1-1/(x+2)`, in the form `y = Q +R/S`, where R is a polynomial of

degree lesser than that of S.

`y = Q` gives a /horizontal/slant/curvilinear asymptote and

{x = a zero of S} gives vertical asymptotes.

Here, `y = Q = 1` gives the horizontal asymptote and

`S = x+2 = 0` gives the vertical asymptote.

Interestingly,

`(y-1)(x+2)=-1` is of the form

`L_1L_2`=constant, where `L_1 and L_2` are linear.

So, this represents a hyperbola with asymptotes

`L_1L_2=(y-1)(x+2)=0`. The asymptotes are at right angles,

and so, the hyperbola is rectangular.

See this with asymptotes, in the Socratic graph.

graph{((y-1)(x+2)+1)(y-1+.001x)(x+2+.001y)=0x^2 [-10, 10, -5, 5]}