Question

The vertical asymptote of the curve $y=\frac{x^2-3x}{x-4}$ is:

• A. $x=4$
• B. $x=3$
• C. $x=0$
• D. Does not exist

Answer 1

The correct option is A $x=4$

Given: $y=\frac{x^2-3x}{x-4}$
Domain is $R-\left\{4\right\}$
$\underset{x\to 4^{+}}{lim}\left(\frac{x^2-3x}{x-4}\right)=\infty \underset{x\to 4^{-}}{lim}\left(\frac{x^2-3x}{x-4}\right)=-\infty$
$\therefore x=4$ is the vertical asymptote.