1. **State the problem:** We are given the function $f(x) = (x^2 - 4)^{12}$ and want to understand its behavior and graph. 2. **Recall the formula and important rules:** The function is a composition of a polynomial inside a power. The base polynomial is $x^2 - 4$, which factors as $(x-2)(x+2)$. 3. **Analyze the base polynomial:** The roots of $x^2 - 4$ are $x = 2$ and $x = -2$. At these points, $f(x) = 0^{12} = 0$. 4. **Effect of the exponent 12:** Since 12 is an even positive integer, the function will be non-negative for all $x$ because any real number raised to an even power is non-negative. 5. **Behavior near the roots:** Near $x = \\pm 2$, the function touches the x-axis but does not cross it because the even power causes the graph to "bounce" off the axis. 6. **Behavior outside the roots:** For $|x| > 2$, $x^2 - 4 > 0$, so $f(x)$ is positive and grows rapidly as $|x|$ increases. 7. **Behavior inside the roots:** For $-2 < x < 2$, $x^2 - 4 < 0$, but raising to the 12th power makes $f(x)$ positive again. 8. **Summary:** The graph has zeros at $x = -2$ and $x = 2$, touches the x-axis at these points, and is positive everywhere else. The shape is symmetric about the y-axis because $f(x)$ depends on $x^2$. Final answer: The function $f(x) = (x^2 - 4)^{12}$ has zeros at $x = -2$ and $x = 2$, is non-negative everywhere, and the graph touches but does not cross the x-axis at these points.