Answer
$y=2x+7$ is Oblique asymptote.
No vertical asymptote
Work Step by Step
$R(x)=\frac{8x^2+26x-7}{4x-1}=\frac{4x-1)(2x+7)}{4x-1}=2x+7,x\not=1/4$
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.
$\begin{array} x &2x+7\\
&-- -- -- --\\
4x-1|& 8x^2+26x-7\\
& -8x^2+2x\\
& -- -- -- -- \\
& 28x-7\\
& -28x+7\\
&-- -- -- -- \\
& 0
\end{array}$
Thus, $y=2x+7$ is Oblique asymptote.
There is no vertical asymptote
