1. **State the problem:** We want to analyze the rational function $$y = \frac{2x^2 - 4}{2x^4 - 5x^2 + 2}$$ to find points of exclusion (POEs), vertical asymptotes (VAs), zeros (x-intercepts), end behavior, and asymptote type. 2. **Factor numerator and denominator:** - Numerator: $$2x^2 - 4 = 2(x^2 - 2)$$ - Denominator: $$2x^4 - 5x^2 + 2$$ can be seen as a quadratic in $x^2$. Let $u = x^2$, then denominator becomes $$2u^2 - 5u + 2$$. 3. **Factor denominator quadratic:** We look for two numbers that multiply to $2 \times 2 = 4$ and add to $-5$. These are $-4$ and $-1$. Rewrite denominator: $$2u^2 - 4u - u + 2 = (2u^2 - 4u) - (u - 2) = 2u(u - 2) -1(u - 2) = (2u - 1)(u - 2)$$ Substitute back $u = x^2$: $$2x^4 - 5x^2 + 2 = (2x^2 - 1)(x^2 - 2)$$ 4. **Simplify the function:** $$y = \frac{2(x^2 - 2)}{(2x^2 - 1)(x^2 - 2)}$$ Cancel common factor $x^2 - 2$: $$y = \frac{\cancel{2}(\cancel{x^2 - 2})}{(2x^2 - 1)(\cancel{x^2 - 2})} = \frac{2}{2x^2 - 1}$$ 5. **Points of Exclusion (POEs):** POEs occur where the canceled factor equals zero: $$x^2 - 2 = 0 \Rightarrow x = \pm \sqrt{2}$$ 6. **Vertical Asymptotes (VAs):** VAs occur where the denominator equals zero and is not canceled: $$2x^2 - 1 = 0 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$ 7. **Zeros (x-intercepts):** Zeros occur where numerator equals zero: $$2 = 0$$ has no solution, so no zeros. 8. **End behavior:** Degree numerator after simplification is 0 (constant 2), degree denominator is 2 (from $2x^2 - 1$). As $x \to \pm \infty$, denominator grows large, so: $$y \to \frac{2}{\infty} = 0$$ So horizontal asymptote is $y=0$. 9. **Asymptote type:** - Vertical asymptotes at $x = \pm \frac{\sqrt{2}}{2}$ - Horizontal asymptote at $y=0$ **Summary:** - POEs: $x = \pm \sqrt{2}$ - VAs: $x = \pm \frac{\sqrt{2}}{2}$ - Zeros: none - End behavior: $y \to 0$ - Asymptotes: vertical at $x=\pm \frac{\sqrt{2}}{2}$, horizontal at $y=0$