How do you find the vertical, horizontal and slant asymptotes of: f(x)=(x+1)/(2x+10)?

Jan 14, 2017

The curve is a rectangular hyperbola.
Horizontal asymptote : $\leftarrow y = \frac{1}{2} \rightarrow$
Vertical asymptote : $\uparrow x = - 5 \downarrow$

Explanation:

An equation in the form

$\left(y - {m}_{1} x - {c}_{1}\right) \left(y - {m}_{2} x - {c}_{2}\right) = c \ne 0$ represents a hyperbola

contained between the asymptotes

$\left(y - {m}_{1} x - {c}_{1}\right) \left(y - {m}_{2} x - {c}_{2}\right) = 0$

Here, cross multiplying and rearranging,

$\left(2 y - 1\right) \left(x + 5\right) = 6$

The asymptotes at right angles are

#2y-1)(x+5)=0, giving

$x = - 5 \mathmr{and} y = \frac{1}{2}$

Now, see the asymptotes-inclusive graph.

graph{((2y-1)(x+5)-6)(2y-1)(x+.000000001y+5)=0 [-15, 5, -7.5, 7.5]}