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

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

divide terms on numerator/denominator by x

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]}

To find the vertical, horizontal, and oblique asymptotes for the function $f(x) = \frac{x+6}{2x+1}$:

Vertical asymptote: Set the denominator equal to zero and solve for $x$. The vertical asymptote occurs where the denominator equals zero but the numerator does not. In this case, $2x + 1 = 0$ gives $x = -\frac{1}{2}$. So, there is a vertical asymptote at $x = -\frac{1}{2}$.

Horizontal asymptote: If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. In this case, the degrees are equal (both 1), so the horizontal asymptote is the ratio of the leading coefficients, which is $\frac{1}{2}$.

Oblique asymptote: To find the oblique asymptote, divide the numerator by the denominator using polynomial long division or synthetic division. In this case, dividing $x+6$ by $2x+1$ yields $\frac{1}{2} + \frac{11}{4(2x+1)}$. The quotient of the division represents the oblique asymptote. So, the oblique asymptote is $y = \frac{1}{2}$.