Question

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

Answer 1

`x = -1/2`
`y = 1/2`

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : 2x + 1 = 0 → 2x = -1 → `x = -1/2" is the asymptote "`

Horizontal asymptotes occur as `lim_(x to +- oo) , f(x) to 0`

divide terms on numerator/denominator by x

`rArr (x/x + 6/x)/((2x)/x + 1/x) = (1 + 6/x)/(2 + 1/x)`

as `x to +- oo , y to (1+0)/(2 + 0)`

`rArr y = 1/2" is the asymptote "`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(x+6)/(2x+1) [-10, 10, -5, 5]}