1. **State the problem:** We are given a function $y=5^x$ with domain $x \geq 0$ and points $(1,4)$ and $(2,16)$ that the graph passes through. 2. **Check if the points lie on the graph $y=5^x$:** - For $x=1$, $y=5^1=5$, but the point is $(1,4)$, so it does not match. - For $x=2$, $y=5^2=25$, but the point is $(2,16)$, so it does not match. 3. **Conclusion:** The points $(1,4)$ and $(2,16)$ do not lie on the graph of $y=5^x$. Therefore, the graph passing through these points is not $y=5^x$. 4. **Find the exponential function passing through the points:** Assume $y=a^x$ passes through $(1,4)$ and $(2,16)$. 5. Use the points to find $a$: $$4 = a^1 = a$$ $$16 = a^2$$ 6. Substitute $a=4$ into the second equation: $$16 = 4^2 = 16$$ This is true, so $a=4$. 7. **Final function:** $$y=4^x$$ **Answer:** The exponential function passing through $(1,4)$ and $(2,16)$ is $y=4^x$, not $y=5^x$.