1. **State the problem:** Graph the function $$f(x) = \frac{2x^4 - x^2 + 1}{x^2 - 4}$$ and analyze its key features. 2. **Identify domain restrictions:** The denominator is $$x^2 - 4 = (x-2)(x+2)$$ which is zero at $$x=2$$ and $$x=-2$$. These points are vertical asymptotes because the function is undefined there. 3. **Analyze end behavior:** For large $$|x|$$, the highest degree terms dominate. The numerator behaves like $$2x^4$$ and the denominator like $$x^2$$. So, $$f(x) \approx \frac{2x^4}{x^2} = 2x^2$$ for large $$|x|$$. This means the function grows approximately like $$2x^2$$ as $$x \to \pm \infty$$. 4. **Find horizontal or oblique asymptotes:** Since the degree of numerator (4) is greater than denominator (2), no horizontal asymptote. Instead, the function behaves like $$2x^2$$ for large $$|x|$$, so the graph will resemble a parabola opening upward far from the origin. 5. **Check for intercepts:** - **y-intercept:** Evaluate $$f(0) = \frac{2(0)^4 - (0)^2 + 1}{0^2 - 4} = \frac{1}{-4} = -\frac{1}{4}$$. - **x-intercepts:** Solve numerator $$2x^4 - x^2 + 1 = 0$$. Let $$y = x^2$$, then $$2y^2 - y + 1 = 0$$. Discriminant $$\Delta = (-1)^2 - 4 \cdot 2 \cdot 1 = 1 - 8 = -7 < 0$$, so no real roots. No x-intercepts. 6. **Summary:** - Vertical asymptotes at $$x = \pm 2$$. - No x-intercepts. - y-intercept at $$\left(0, -\frac{1}{4}\right)$$. - End behavior like $$2x^2$$. This describes the graph's main features.