Question

Find the asymptotes of the curve \[y=\dfrac{3x+3}{x-3}\].

Answer 1

Hint:An asymptote is a line which touches the curve at infinity or we can say that it is a line such that the distance between the line and the curve approaches zero if the coordinates tend to infinity there will be two asymptotes of the given function one is horizontal and the other one is vertical. When we take denominator as equal to zero, we will get vertical asymptotes and if we take the \[\underset{x\to \infty }{\mathop{\lim }}\,\] for the function then we get horizontal asymptotes.

Complete step-by-step answer:
We have been given the curve \[y=\dfrac{3x+3}{x-3}\].
We know that there will be two asymptotes one is horizontal and the other one is vertical. By denominator is equal to zero, we get the vertical asymptotes and if we take \[x\to \infty \] limit on the function, we get horizontal asymptotes.
For the vertical asymptotes, we have,
\[\begin{align}
  & x-3=0 \\
 & \Rightarrow x=3 \\
\end{align}\]
Hence the vertical asymptotes is x=3.
Now for the horizontal asymptotes we have,
\[\Rightarrow Y=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{3x+3}{x-3} \right)\]
On taking ‘x’ as common, we get as follows:
\[\begin{align}
  & \Rightarrow Y=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{x\left( 3+\dfrac{3}{x} \right)}{x\left( 1-\dfrac{3}{x} \right)} \\
 & \Rightarrow Y=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{3+\dfrac{3}{x}}{\left( 1-\dfrac{3}{x} \right)} \\
 & \Rightarrow Y=3 \\
\end{align}\]
Hence the horizontal asymptote is Y=3.
Therefore, we get the asymptotes of the given function are x=3 and y=3.

Note: Sometimes we just find the vertical asymptotes and we forget about the horizontal asymptotes. So be careful while finding the horizontal asymptotes as there is a chance of calculation mistake. The graph for this question is as shown below.