함수의 그래프 그리기

Vertical asymptotes occur where the function blows up. In rational functions, this can only happen when the denominator goes to 0. In our case, the denominator is 0 when $x=3$, and in this case the numerator is $147$. That means that as $x$ gets closer and closer to 3, the numerator gets closer and closer to 147 while the denominator gets closer and closer to 0, so $|f(x)|$ grows without bound. That is, there is a vertical asymptote at $x=3$.

The horizontal asymptotes are found by taking the limits as $x$ goes to infinity and negative infinity. In our case, they are the same, so we condense our work.

$$\begin{aligned} \lim_{x \to \pm \infty}\dfrac{x(2x+1)(x-7)}{3x^3-81} &=\lim_{x \to \pm \infty}\dfrac{2x^3+ax^2+bx+c}{3x^3-81} \\ \text{where $a$, $b$, ad $c$ are some constants. Remember, for rational functions, you can figure out the end behaviour by looking only at the terms with the highest degree--the others won't matter, so we don't bother finding them. From here, we divide the numerator and denominator by the highest power of $x$ in the denominator, $x^3$.} \\ &=\lim_{x \to \pm \infty}\dfrac{2x^3+ax^2+bx+c}{3x^3-81}\left(\frac{\tfrac{1}{x^3}}{\tfrac{1}{x^3}}\right)\\ &=\lim_{x \to \pm \infty}\dfrac{2+\tfrac{a}{x}+\tfrac{b}{x^2}+\tfrac{c}{x^3}}{3-\tfrac{81}{x^3}}\\ &=\dfrac{2+0+0+0}{3-0}=\frac{2}{3} \end{aligned}$$

So there is a horizontal asymptote of $y=\dfrac{2}{3}$ both as $x \to \infty$ and as $x \to -\infty$.