What are the vertical and horizontal asymptotes of the rational function \( f(x)=\frac{2 x^{2}-1}{x^{2}-16} ? \)

To determine the vertical and horizontal asymptotes of the rational function \( f(x) = \frac{2x^{2} - 1}{x^{2} - 16} \), follow these steps: ### 1. **Vertical Asymptotes** Vertical asymptotes occur where the denominator is zero, provided the numerator is not also zero at those points. **Step-by-Step:** - **Set the denominator equal to zero:** \[ x^{2} - 16 = 0 \implies x^{2} = 16 \implies x = \pm 4 \] - **Check the numerator at these values:** \[ f(4) = \frac{2(4)^2 - 1}{4^2 - 16} = \frac{32 - 1}{16 - 16} = \frac{31}{0} \quad (\text{undefined}) \] \[ f(-4) = \frac{2(-4)^2 - 1}{(-4)^2 - 16} = \frac{32 - 1}{16 - 16} = \frac{31}{0} \quad (\text{undefined}) \] Since the numerator is not zero at \( x = 4 \) and \( x = -4 \), these are indeed vertical asymptotes. **Vertical Asymptotes:** \[ x = 4 \quad \text{and} \quad x = -4 \] ### 2. **Horizontal Asymptote** Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator. - **Degrees of the polynomials:** - Numerator: Degree 2 - Denominator: Degree 2 Since the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. **Step-by-Step:** - **Leading coefficient of the numerator:** 2 - **Leading coefficient of the denominator:** 1 **Horizontal Asymptote:** \[ y = \frac{2}{1} = 2 \] ### **Summary** - **Vertical Asymptotes:** \( x = 4 \) and \( x = -4 \) - **Horizontal Asymptote:** \( y = 2 \) **Graphical Representation:**  *(Note: Replace the image link with an actual graph if needed.)* **Final Answer:** Vertical asymptotes at x = 4 and x = –4, and a horizontal asymptote of y = 2.