Question

How do you find vertical, horizontal and oblique asymptotes for `(6x + 6) / (3x^2 + 1)`?

Answer 1

`"horizontal asymptote at "y=0`

Explanation:

`"let "f(x)=(6x+6)/(3x^2+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 "3x^2+1=0rArrx^2=-1/3`

`"This has no real solutions hence there are no vertical"`
`"asymptotes"`

`"Horizontal asymptotes occur as"`

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

`"divide terms on numerator/denominator by the highest"`
`"power of "x" that is "x^2`

`f(x)=((6x)/x^2+6/x^2)/((3x^2)/x^2+1/x^2)=(6/x+6/x^2)/(3+1/x^2)`

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

`y=0" is the asymptote"`

`"Oblique asymptotes occur when the degree of the"`
`"numerator is greater than the degree of the denominator. "`
`"This is not the case here hence no oblique asymptotes"`

graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}