1. **State the problem:** Find the points of exclusion, vertical asymptotes, zeros (x-intercepts), factor the expression, and analyze the end behavior and asymptote type for the function $$y = \frac{2x^2 - 4}{2x^4 - 5x^2 + 2}$$ 2. **Points of exclusion:** These occur where the denominator is zero because the function is undefined there. Set denominator equal to zero: $$2x^4 - 5x^2 + 2 = 0$$ Let $u = x^2$, then: $$2u^2 - 5u + 2 = 0$$ 3. **Solve quadratic equation:** Use 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, $$u_1 = \frac{5 + 3}{4} = 2$$ $$u_2 = \frac{5 - 3}{4} = \frac{1}{2}$$ 4. **Back-substitute $u = x^2$:** $$x^2 = 2 \implies x = \pm \sqrt{2}$$ $$x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$ These are the points of exclusion. 5. **Vertical asymptotes:** Since the denominator is zero at these points and numerator is not zero there, vertical asymptotes occur at $$x = \pm \sqrt{2}, \quad x = \pm \frac{\sqrt{2}}{2}$$ 6. **Zeros (x-intercepts):** Set numerator equal to zero: $$2x^2 - 4 = 0 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \pm \sqrt{2}$$ 7. **Check if zeros are also points of exclusion:** Since $x=\pm \sqrt{2}$ are zeros of numerator and denominator, these are removable discontinuities (holes), not vertical asymptotes. 8. **Factor numerator and denominator:** Numerator: $$2x^2 - 4 = 2(x^2 - 2) = 2(x - \sqrt{2})(x + \sqrt{2})$$ Denominator: $$2x^4 - 5x^2 + 2 = (2x^2 - 4)(x^2 - \frac{1}{2})$$ Check factorization by quadratic roots: $$2x^4 - 5x^2 + 2 = (x^2 - 2)(2x^2 - 1)$$ So denominator factors as: $$ (x - \sqrt{2})(x + \sqrt{2})(\sqrt{2}x - 1)(\sqrt{2}x + 1)$$ 9. **Simplify the function by canceling common factors:** $$y = \frac{2(x - \sqrt{2})(x + \sqrt{2})}{(x - \sqrt{2})(x + \sqrt{2})(\sqrt{2}x - 1)(\sqrt{2}x + 1)}$$ Cancel common factors: $$y = \frac{\cancel{2}(\cancel{x - \sqrt{2}})(\cancel{x + \sqrt{2}})}{\cancel{(x - \sqrt{2})}(\cancel{x + \sqrt{2}})(\sqrt{2}x - 1)(\sqrt{2}x + 1)} = \frac{2}{(\sqrt{2}x - 1)(\sqrt{2}x + 1)}$$ 10. **End behavior:** For large $x$, denominator behaves like $$(\sqrt{2}x)(\sqrt{2}x) = 2x^2$$ So, $$y \approx \frac{2}{2x^2} = \frac{1}{x^2} \to 0$$ as $x \to \pm \infty$. Horizontal asymptote is $y=0$. 11. **Asymptote type:** - Vertical asymptotes at $x = \pm \frac{\sqrt{2}}{2}$ - Holes (removable discontinuities) at $x = \pm \sqrt{2}$ - Horizontal asymptote at $y=0$ **Final answers:** - Points of exclusion: $x = \pm \sqrt{2}, \pm \frac{\sqrt{2}}{2}$ - Vertical asymptotes: $x = \pm \frac{\sqrt{2}}{2}$ - Zeros (x-intercepts): None (since zeros coincide with holes) - End behavior: $y \to 0$ as $x \to \pm \infty$ - Asymptote type: Vertical asymptotes and horizontal asymptote at $y=0$