How do you find the asymptotes for $y=(x-3)/(2x+5)$ ?

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Answer 1 · 2016-03-20T07:15:45

Vertical Asymptote: $x=-5/2$
Horizontal Asymptote: $y=1/2$

For the vertical asymptote:

Equate the denominator to zero, then solve for x
$2x+5=0$
$2x=-5$
$x=-5/2$

For the horizontal asymptote:

Take the limit of the function

$lim_(xrarr oo) y=lim_(xrarr oo) (x-3)/(2x+5)=1/2$

and therefore

$y=1/2$ is a horizontal asymptote

graph{(y- (x-3)/(2x+5))(y-1/2)=0[-20,20,-10,10]}

God bless....I hope the explanation is useful.

How do you find the asymptotes for y=(x-3)/(2x+5)y=(x-3)/(2x+5)?