1. **State the problem:** We are given the function $$y = (x^2 + x^{-2})^2$$ and want to understand its behavior and graph. 2. **Recall the formula and rules:** The function involves powers of $x$ and their inverses. Remember that $x^{-2} = \frac{1}{x^2}$, so the expression inside the parentheses is $$x^2 + \frac{1}{x^2}$$. 3. **Expand the square:** Using the formula $$(a+b)^2 = a^2 + 2ab + b^2$$, we get $$y = (x^2)^2 + 2 \cdot x^2 \cdot x^{-2} + (x^{-2})^2 = x^4 + 2 + x^{-4}$$. 4. **Simplify the expression:** Since $x^{-4} = \frac{1}{x^4}$, the function can be written as $$y = x^4 + 2 + \frac{1}{x^4}$$. 5. **Analyze the function:** - The term $x^4$ grows very large as $|x|$ increases. - The term $\frac{1}{x^4}$ grows very large as $x$ approaches zero. - The function is always greater than or equal to 2 because $x^4 + \frac{1}{x^4} \geq 2$ by AM-GM inequality. 6. **Symmetry:** The function is even since all powers of $x$ are even, so the graph is symmetric about the y-axis. 7. **Behavior near zero and infinity:** - As $x \to 0$, $y \to \infty$ due to $\frac{1}{x^4}$. - As $x \to \pm \infty$, $y \to \infty$ due to $x^4$. 8. **Minimum value:** The minimum occurs when $x^4 + \frac{1}{x^4}$ is minimized, which is at $x^4 = 1$, so $x = \pm 1$. At these points, $$y = 1 + 2 + 1 = 4$$. **Final answer:** The function is $$y = x^4 + 2 + \frac{1}{x^4}$$ with minimum value 4 at $x = \pm 1$, symmetric about the y-axis, and tends to infinity as $x$ approaches 0 or infinity.