1. Let's analyze the behavior of the function as $x$ approaches negative infinity. 2. Suppose the function is $y = a^x$ where $a > 0$ and $a \neq 1$. 3. If $a > 1$, then as $x \to -\infty$, $a^x = \frac{1}{a^{-x}}$ and since $-x \to +\infty$, $a^{-x} \to +\infty$, so $a^x \to 0$. 4. This means $y$ approaches zero, not positive infinity, as $x$ goes to negative infinity. 5. If $0 < a < 1$, then as $x \to -\infty$, $a^x$ behaves like $\left(\frac{1}{b}\right)^x$ where $b = \frac{1}{a} > 1$. 6. So $a^x = b^{-x}$ and since $-x \to +\infty$, $b^{-x} \to 0$, so again $y \to 0$. 7. Therefore, for any valid exponential function $y = a^x$ with $a > 0$ and $a \neq 1$, as $x \to -\infty$, $y \to 0$. 8. If you observe $y$ going to positive infinity as $x \to -\infty$, the function might be different or there might be a misunderstanding. 9. Please check the function definition or provide it for further clarification.