1. **State the problem:** We are given the function $y=2^{x^2-4}$ and want to understand its behavior. 2. **Formula and rules:** The function is an exponential function with base 2 and exponent $x^2-4$. Recall that exponential functions of the form $y=a^{f(x)}$ grow or decay depending on $f(x)$. 3. **Analyze the exponent:** The exponent is $x^2-4$. This is a quadratic expression which is always greater or equal to $-4$ since $x^2 \geq 0$. 4. **Evaluate key points:** - At $x=0$, $y=2^{0^2-4}=2^{-4}=\frac{1}{2^4}=\frac{1}{16}$. - At $x=\pm 2$, $y=2^{(\pm 2)^2-4}=2^{4-4}=2^0=1$. 5. **Behavior:** As $|x|$ increases beyond 2, $x^2-4$ becomes positive and $y$ grows exponentially. 6. **Summary:** The function has a minimum value at $x=0$ of $\frac{1}{16}$ and grows exponentially as $|x|$ increases beyond 2. Final answer: The function is $y=2^{x^2-4}$ with minimum $\frac{1}{16}$ at $x=0$ and grows exponentially for large $|x|$.