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]}

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