The graph of the rational function $\frac{2x^6+3x^5 - x^2 - 1}{q(x)}$  has a horizontal asymptote. What is the smallest possible degree of $q(x)$?
For the given function to have a horizontal asymptote, the function will need to approach a constant as $x \to \pm \infty$. This is only possible if the denominator $q(x)$ is at least the same degree as the numerator. Since the numerator has degree $6$, the smallest possible degree of $q(x)$ that will allow the function to have a horizontal asymptote is $\boxed{6}$.  For example, we can take $q(x) = x^6.$
6
Level 3
Intermediate Algebra
Problem: The graph of the rational function $\frac{2x^6+3x^5 - x^2 - 1}{q(x)}$  has a horizontal asymptote. What is the smallest possible degree of $q(x)$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the smallest possible degree of the polynomial $q(x)$ such that the rational function $\frac{2x^6+3x^5 - x^2 - 1}{q(x)}$ has a horizontal asymptote, we need to analyze the degrees of the numerator and denominator. The degree of the numerator, $2x^6 + 3x^5 - x^2 - 1$, is $6$ because the highest power of $x$ is $6$. 

For a rational function to have a horizontal asymptote, the degree of the numerator must be less than or equal to the degree of the denominator. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is determined by the ratio of the leading coefficients. 

In this case, to ensure that the function has a horizontal asymptote, the degree of $q(x)$ must be at least $6$. If the degree of $q(x)$ is less than $6$, the function will not have a horizontal asymptote, as the numerator will dominate the behavior of the function as $x$ approaches infinity. 

Thus, the smallest degree of $q(x)$ that allows for a horizontal asymptote is $6$.<|END_OF_LATENT|> The final answer is $6$.