AP Calculus BC Question of the Day

1. $y=2$
2. $x=2$
3. $y=0$ (correct answer)
4. $x=\sqrt[3]{8}$
5. $y=\dfrac{1}{2}$

Explanation: This problem requires finding the horizontal asymptote of $q(x)=\frac{2x^2+1}{x^3-8}$ as $x\to\infty$. The numerator has degree 2 while the denominator has degree 3, so the denominator's degree is higher. When the denominator has higher degree than the numerator, the horizontal asymptote is always $y=0$. As $x$ becomes very large, the denominator $x^3-8$ grows much faster than the numerator $2x^2+1$, causing the fraction to approach 0. Choice D ($x=\sqrt[3]{8}$) might confuse students who are thinking about where the denominator equals zero (a vertical asymptote issue, not horizontal). For rational functions where the bottom degree exceeds the top degree, the function always approaches zero as $x\to\pm\infty$.