Question

How do you find the asymptotes for `(3x-2) / (x+1)`?

Answer 1

When `x=-1 =>-5/0` so there is an asymptote when `x=-1`

Answer 2

`"vertical asymptote at "x=-1`
`"horizontal asymptote at "y=3`

Explanation:

`"let "f(x)=(3x-2)/(x+1)`

The denominator of f(x) cannot be zero as this would make f(x) 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+1=0rArrx=-1" is the asymptote"`

`"horizontal asymptotes occur as"`

`lim_(xto+-oo),f(x)toc" (a constant)"`

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

`f(x)=((3x)/x-2/x)/(x/x+1/x)=(3-2/x)/(1+1/x)`

`"as "xto+-oo,f(x)to(3-0)/(1+0)`

`rArry=3" is the asymptote"`
graph{(3x-2)/(x+1) [-10, 10, -5, 5]}