12. Find the horizontal and vertical asymptotes. (Do not sketch) \( f(x)=\frac{-2 x^{2}+4 x+1}{9 x^{2}-1} \)

1. **Horizontal Asymptote** Since the degrees of the numerator and the denominator are equal (both are degree 2), the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is \(-2\) and of the denominator is \(9\). Thus, the horizontal asymptote is \[ y = \frac{-2}{9}. \] 2. **Vertical Asymptotes** To find the vertical asymptotes, set the denominator equal to zero and solve: \[ 9x^2 - 1 = 0. \] Factor the expression: \[ (3x - 1)(3x + 1) = 0. \] Set each factor equal to zero: \[ 3x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{3}, \] \[ 3x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{3}. \] Check that these values do not also cancel a factor in the numerator. Evaluating the numerator at \( x = \frac{1}{3} \) and \( x = -\frac{1}{3} \) shows that the numerator does not equal zero at these points, confirming that they are vertical asymptotes. Therefore, the vertical asymptotes are \[ x = \frac{1}{3} \quad \text{and} \quad x = -\frac{1}{3}. \] **Final Answer:** - Horizontal Asymptote: \( y = -\frac{2}{9} \) - Vertical Asymptotes: \( x = \frac{1}{3} \) and \( x = -\frac{1}{3} \)