Answer
$y=x+5$ is Oblique asymptote.
$x\in \{-2, 7\}$ is a vertical asymptote
Work Step by Step
$G(x)=\frac{x^3+1}{x^2-5x+14},$
The degree of the leading coefficient of the numerator is, $n=3$. the degree of the leading coefficient the denominator is, $m=2$.
Thus, $n=m+1,$
When $n=m+1,$ the quotient of the division of numerator by denominator is Oblique asymptote.
$\begin{array} x &x+5\\
&-- -- -- --\\
x^2-5x+14|& x^3+1\\
& -x^3+5x^2+14x\\
& -- -- -- -- \\
& 5x^2+14x+1\\
& -5x^2+25x-70\\
&-- -- -- -- \\
& 39x-69
\end{array}$
Thus, $y=x+5$ is Oblique asymptote.
Factorizing the denominator,
$x^2-5x+14=x^2+2x-7x-14,$
$=x(x+2)-7(x+2),$
$=(x-7)(x+2),$
Thus, $x\in \{-2, 7\}$ is a vertical asymptote
