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

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Answer 1 · 2017-02-20T03:44:59

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

#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]}