1. The problem asks to analyze the function $f(x) = 16^{9-x}$. 2. This is an exponential function where the base is 16 and the exponent is $9-x$. 3. Recall the exponential function formula: $f(x) = a^{bx+c}$ where $a>0$ and $a \neq 1$. 4. Here, $a=16$, $b=-1$, and $c=9$. 5. The function decreases because the exponent has a negative coefficient for $x$. 6. To understand the behavior, rewrite the function as: $$f(x) = 16^{9-x} = 16^9 \cdot 16^{-x} = 16^9 \cdot \left(\frac{1}{16}\right)^x$$ 7. This shows the function is a decreasing exponential function scaled by $16^9$. 8. The domain of $f(x)$ is all real numbers $(-\infty, \infty)$. 9. The range is $(0, \infty)$ because exponential functions with positive bases are always positive. 10. The function has no intercepts with the x-axis because it never equals zero. 11. The y-intercept is at $x=0$: $$f(0) = 16^{9-0} = 16^9$$ 12. The graph will approach zero as $x \to \infty$ and grow very large as $x \to -\infty$. Final answer: The function $f(x) = 16^{9-x}$ is a decreasing exponential function with domain $(-\infty, \infty)$ and range $(0, \infty)$, y-intercept at $(0, 16^9)$, and no x-intercepts.