Question

find the slant asymptote and the vertical asymptotes. (if an answer does not exist, enter dne )
$r(x) = \frac{x^3 + 4}{2x^2 + x - 1}$
slant asymptote
vertical asymptote (smaller value)
vertical asymptote (larger value)
sketch a graph of the function.

Explanation:

Step1: Find vertical asymptotes

Factor denominator: $2x^2 + x - 1 = (2x - 1)(x + 1)$. Set to zero: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$; $x + 1 = 0 \Rightarrow x = -1$.

Step2: Find slant asymptote

Divide numerator by denominator: $x^3 + 4 = (2x^2 + x - 1)(\frac{1}{2}x - \frac{1}{4}) + \frac{5}{4}x + \frac{15}{4}$. The quotient is $\frac{1}{2}x - \frac{1}{4}$.

Answer:

slant asymptote: $y = \frac{1}{2}x - \frac{1}{4}$
vertical asymptote (smaller value): $x = -1$
vertical asymptote (larger value): $x = \frac{1}{2}$