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

1 Answer
Dec 6, 2015

Answer:

Horizontal asymptotes are #y=1/2# at both positive and negative infinite. Furthermore, we have that #x=-1/2# is a vertical asymptote.

Explanation:

Horizontal asymptotes: we must look for

#lim_{x\to\pm\infty} f(x)#

Both limits are #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=-1/2#, because for that value the denominator is zero. We have that the numerator at #x=-1/2# is positive.

So, if #x\to\(-1/2)^-#, the denominator approaches zero from negative values, and so we have negative infinity.
On the opposite, if #x\to\(-1/2)^+#, 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.