Question

1. (04.03 mc) find the horizontal asymptote of f(x) = $\frac{-3x^{2}+3x + 6}{x^{2}+1}$. (5 points) y = -1 y = -3 y = 1 y = 3

Explanation:

Step1: Identify degrees of polynomials

The degree of the numerator $-3x^{2}+3x + 6$ is $n = 2$, and the degree of the denominator $x^{2}+1$ is $m=2$.

Step2: Use horizontal - asymptote rule

When $n = m$, the horizontal asymptote $y$ is the ratio of the leading - coefficients. The leading coefficient of the numerator is $a=-3$ and the leading coefficient of the denominator is $b = 1$. So $y=\frac{a}{b}=\frac{-3}{1}=-3$.

Answer:

B. $y = - 3$