1. **State the problem:** We need to analyze the function $$y = \frac{2x^2 - 4}{2x^4 - 5x^2 + 2}$$ by factoring it, finding points of exclusion (POE), vertical asymptotes (VA), zeros (x-intercepts), end behavior, and asymptote type. 2. **Factor numerator and denominator:** - Numerator: $$2x^2 - 4 = 2(x^2 - 2)$$ - Denominator: $$2x^4 - 5x^2 + 2$$. Let $$u = x^2$$, then denominator becomes $$2u^2 - 5u + 2$$. 3. **Factor denominator as quadratic in $$u$$:** Use the quadratic formula: $$u = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$$ So, $$u_1 = \frac{5 + 3}{4} = 2$$ $$u_2 = \frac{5 - 3}{4} = \frac{1}{2}$$ Therefore, $$2x^4 - 5x^2 + 2 = 2(x^2 - 2)(x^2 - \frac{1}{2})$$ 4. **Simplify the function:** $$y = \frac{2(x^2 - 2)}{2(x^2 - 2)(x^2 - \frac{1}{2})}$$ Cancel common factor $x^2 - 2$: $$y = \frac{\cancel{2}(\cancel{x^2 - 2})}{\cancel{2}(\cancel{x^2 - 2})(x^2 - \frac{1}{2})} = \frac{1}{x^2 - \frac{1}{2}}$$ 5. **Points of Exclusion (POE):** The canceled factor $x^2 - 2 = 0$ gives POE: $$x^2 = 2 \Rightarrow x = \pm \sqrt{2}$$ 6. **Vertical Asymptotes (VA):** Set denominator of simplified function to zero: $$x^2 - \frac{1}{2} = 0 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$ 7. **Zeros (x-intercepts):** Set numerator of simplified function to zero. Numerator is 1, so no zeros. 8. **End behavior:** As $$x \to \pm \infty$$, the dominant term in denominator is $$x^2$$, so $$y \approx \frac{1}{x^2} \to 0^+$$ 9. **Asymptote type:** Horizontal asymptote at $$y = 0$$. **Final answers:** - POE: $$x = \pm \sqrt{2}$$ - VA: $$x = \pm \frac{\sqrt{2}}{2}$$ - Zeros: none - End behavior: $$y \to 0^+$$ as $$x \to \pm \infty$$ - Asymptote type: horizontal asymptote at $$y=0$$