1. **Problem statement:** Describe the behavior of the exponential function $K(n) = K_0 \cdot q^n$ for $K_0 > 0$ and $q > 1$. 2. **Formula:** The general formula for exponential growth or decay is: $$K(n) = K_0 \cdot q^n$$ where: - $K_0$ is the initial value (capital at time 0), - $q$ is the growth factor per time unit, - $n$ is the time or input variable. 3. **Important rules:** - If $q > 1$, the function grows exponentially. - If $0 < q < 1$, the function decays exponentially. - If $q = 1$, the function is constant. - If $q = 0$, the function is zero for all $n > 0$. 4. **Given:** $K_0 = 1$ and $q = 1.03$ (3% growth rate). 5. **Behavior:** - At $n=0$, $K(0) = 1 \cdot 1.03^0 = 1$. - For increasing $n$, $K(n)$ increases because $q > 1$. - The function grows without bound as $n \to \infty$. 6. **Interpretation:** - The capital grows each year by 3% compounded. - This is the classic compound interest effect. 7. **Summary:** - The graph crosses the y-axis at $K_0=1$. - It rises exponentially to positive infinity as $n$ increases. **Final answer:** $$K(n) = 1 \cdot 1.03^n$$ with $K(n)$ growing exponentially for $n \geq 0$.