Find the domain and the vertical and horizontal asymptotes. Separate your asymptote answers with commas, if there are multiple, or enter "DNE" if there are none. \( f(x)=\frac{-x^{2}+9}{-3 x^{2}+12} \) Domain: Vertical asymptote(s): \( x=\square \) Horizontal asymptote(s): \( y=\square \)

To find the domain, vertical asymptotes, and horizontal asymptotes of the function \( f(x) = \frac{-x^{2}+9}{-3x^{2}+12} \), we will analyze the function step by step. ### Domain The domain of a rational function is all real numbers except where the denominator is zero. 1. Set the denominator equal to zero: \[ -3x^{2} + 12 = 0 \] \[ -3x^{2} = -12 \] \[ x^{2} = 4 \] \[ x = 2 \quad \text{or} \quad x = -2 \] Thus, the function is undefined at \( x = 2 \) and \( x = -2 \). Therefore, the domain is: \[ \text{Domain: } (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \] ### Vertical Asymptotes Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. From our previous calculation, the vertical asymptotes are at \( x = 2 \) and \( x = -2 \). We check the numerator at these points: - For \( x = 2 \): \[ -x^{2} + 9 = -2^{2} + 9 = -4 + 9 = 5 \quad (\text{not zero}) \] - For \( x = -2 \): \[ -(-2)^{2} + 9 = -4 + 9 = 5 \quad (\text{not zero}) \] Since the numerator is not zero at both points, we confirm the vertical asymptotes: \[ \text{Vertical asymptote(s): } x = 2, x = -2 \] ### Horizontal Asymptotes To find horizontal asymptotes, we compare the degrees of the numerator and denominator. - The degree of the numerator \( -x^{2} + 9 \) is 2. - The degree of the denominator \( -3x^{2} + 12 \) is also 2. Since the degrees are equal, we find the horizontal asymptote by taking the ratio of the leading coefficients: \[ \text{Leading coefficient of the numerator: } -1 \] \[ \text{Leading coefficient of the denominator: } -3 \] \[ \text{Horizontal asymptote: } y = \frac{-1}{-3} = \frac{1}{3} \] ### Summary Putting it all together, we have: - Domain: \( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \) - Vertical asymptote(s): \( x = 2, x = -2 \) - Horizontal asymptote(s): \( y = \frac{1}{3} \) Thus, the final answers are: - Domain: \( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \) - Vertical asymptote(s): \( x = 2, x = -2 \) - Horizontal asymptote(s): \( y = \frac{1}{3} \)