Question

find the slant asymptote and the vertical asymptotes. (if an answer does not exist, enter dne.)
\\( r(x) = \frac{4x^3 + 4x^2}{x^2 - 9} \\)
slant asymptote
vertical asymptote (smaller value)
vertical asymptote (larger value)

Explanation:

Step1: Find vertical asymptotes

Set denominator to zero: $x^2 - 9 = 0 \Rightarrow (x-3)(x+3)=0 \Rightarrow x=\pm3$

Step2: Find slant asymptote

Divide numerator by denominator: $4x^3 + 4x^2 = (4x + 4)(x^2 - 9) + 36x + 36$. As $x\to\pm\infty$, remainder term vanishes, so slant asymptote is $y=4x+4$

Answer:

slant asymptote: $y=4x+4$
vertical asymptote (smaller value): $x=-3$
vertical asymptote (larger value): $x=3$