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

Feb 20, 2017

Vertical : $\uparrow x = - 2 \downarrow$.Horizontal : $\leftarrow y = 1 \rightarrow$. See this asymptotes-inclusive rectangular hyperbola in the Socratic graph.

#### Explanation:

$y = f \left(x\right) = 1 - \frac{1}{x + 2}$, in the form $y = Q + \frac{R}{S}$, 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,

$\left(y - 1\right) \left(x + 2\right) = - 1$ is of the form

${L}_{1} {L}_{2}$=constant, where ${L}_{1} \mathmr{and} {L}_{2}$ are linear.

So, this represents a hyperbola with asymptotes

${L}_{1} {L}_{2} = \left(y - 1\right) \left(x + 2\right) = 0$. 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]}