Question

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

Answer 1

`"vertical asymptote at "x=-2`
`"horizontal asymptote at "y=0`

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

`"solve "x+2=0rArrx=-2" is the asymptote"`

`"Horizontal asymptotes occur as"`

`lim_(xto+-oo),ytoc" ( a constant )"`

`"divide terms on numerator/denominator by "x`

`y=(3/x)/(x/x+2/x)=(3/x)/(1+2/x)`

`"as "xto+-oo,yto0/(1+0)`

`y=0" os the asymptote"`
graph{3/(x+2) [-10, 10, -5, 5]}