How do you find the asymptotes for f(x) = (x+6)/ (2x+1)?

Dec 6, 2015

Horizontal asymptotes are $y = \frac{1}{2}$ at both positive and negative infinite. Furthermore, we have that $x = - \frac{1}{2}$ is a vertical asymptote.

Explanation:

Horizontal asymptotes: we must look for

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right)$

Both limits are $\frac{1}{2}$, because we have the ratio of two polynomials of the same degree, and so the limit is the ratio of the leading coefficient, which is $1$ for the numerator and $2$ for the denominator (they are the coefficient of the $x$ term, which is the higher power of $x$ appearing).

Vertical asymptotes: The only point in which the function is not defined is $x = - \frac{1}{2}$, because for that value the denominator is zero. We have that the numerator at $x = - \frac{1}{2}$ is positive.

So, if $x \setminus \to \setminus {\left(- \frac{1}{2}\right)}^{-}$, the denominator approaches zero from negative values, and so we have negative infinity.
On the opposite, if $x \setminus \to \setminus {\left(- \frac{1}{2}\right)}^{+}$, the denominator approaches zero from positive values, and so we have positive infinity.

Since we have horizontal asymptotes at infinity in both directions, we have no slant asymptotes.