The vertical, horizontal and oblique asymptotes of the rational function R(x) = frac{6x^{2}+19x-7}{3x-1}.

Given:
The rational function is defined by $R(x)=\frac{6x^2+19x-7}{3x-1}$.
Result used:
Vertical asymptote:
Consider a rational function $R(x)=\frac{p(x)}{q(x)}$, in its lowest terms. If ris a real zero of the polynomial q(x), then x = r is a vertical asymptote of the function R.
Horizontal or oblique asymptotes:
Consider a rational function $R(x)=p\frac{x}{q}(x)=\frac{a_n{x}^n+1_{n-1}{x}^{n-1}+...a_{1}x+a_0}{b_m}{x}^{+}{b}_{m-1}{x}^{m-1}+...b_{1}x+b_0$ where nis the degree of the polynomial p(x) and mis the degree of the polynomial q(x). If $n=m+1$, the rational function R has no horizontal asymptote and only has an oblique asymptote given by $y=ax+b$ where $y=ax+b$ is quotient obtained by the polynomial division $\frac{p(x)}{q(x)}$.
Calculation:
Rewrite the rational function $R(x)=\frac{6x^2+19x-7}{3x-1}$ as follows.
$R(x)=\frac{6x^2+19x-7}{3x-1}$
$=\frac{6x^2+21x-2-7}{3x-1}$
$=3x(2x+7)-1\frac{2x+7}{3x-1}$
$=\frac{(2x+7)(3x-1)}{3x-1}$
$=2x+7$
Hence, the rational function $R(x)=\frac{6x^2+19x-7}{3x-1}$ in its lowest terms is $R(x)=2x+7$.
The polynomial in the denominator of $R(x)=2x+7$ is 1 and 1 has no zeros.
Therefore, the rational function $R(x)=\frac{6x^2+19x-7}{3x-1}$ has no vertical asymptote.
Note that, the degree of the polynomial in the numerator is 2 and the degree of the polynomial in the denominator is 1.
That is, the degree of the polynomial in the numerator is 1 more than the degree of the polynomial in the denominator.
Therefore, the rational function $R(x)=\frac{6x^2+19x-7}{3x-1}$ has no horizontal asymptote.
The polynomial division $\frac{6x^2+19x-7}{3x-1}$ gives the quotient as 2x +7.
Therefore, the rational function $\frac{6x^2+19x-7}{3x-1}$ has an oblique asymptote given by $y=2x+7$.
Hence, the rational function $\frac{6x^2+19x-7}{3x-1}$ has no vertical asymptote, no horizontal ext and oblique asymptote is $y=2x$.