1. **State the problem:** We have the function $f(x) = \log_{\frac{1}{3}} x$ with point $A$ as the x-intercept and point $(3, t)$ on the graph. We need to find $t$, coordinates of $A$, the inverse function $f^{-1}$, its asymptote, graph details, and analyze a translation of $f^{-1}$. 2. **Calculate $t$ (4.1):** Since $(3, t)$ lies on $f$, we have $$t = \log_{\frac{1}{3}} 3.$$ Recall that $\log_a b = c$ means $a^c = b$. Here, $$\left(\frac{1}{3}\right)^t = 3.$$ Rewrite $\frac{1}{3} = 3^{-1}$, so $$3^{-t} = 3^1 \implies -t = 1 \implies t = -1.$$ 3. **Coordinates of $A$ (4.2):** The x-intercept is where $f(x) = 0$, so $$\log_{\frac{1}{3}} x = 0 \implies x = \left(\frac{1}{3}\right)^0 = 1.$$ Thus, $A = (1, 0)$. 4. **Equation of $f^{-1}$ (4.3):** The inverse of $f(x) = \log_{\frac{1}{3}} x$ is $$y = \left(\frac{1}{3}\right)^x.$$ This is because the inverse of a logarithm base $a$ is the exponential function base $a$. 5. **Asymptote of $f^{-1}$ (4.4):** Since $f^{-1}(x) = \left(\frac{1}{3}\right)^x$ is an exponential function, its horizontal asymptote is $$y = 0.$$ 6. **Graph details of $f^{-1}$ (4.5):** - Intercepts: - When $x=0$, $y = \left(\frac{1}{3}\right)^0 = 1$ (y-intercept at $(0,1)$). - When $y=0$, no x-intercept since exponential never touches $y=0$. - Another point: - Use $x=3$, then $y = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$. - Asymptote: $y=0$. 7. **Translation of $f^{-1}$ to get $h$ (4.6):** The graph of $h$ is $f^{-1}$ shifted 5 units right: $$h(x) = f^{-1}(x-5) = \left(\frac{1}{3}\right)^{x-5} = \frac{\left(\frac{1}{3}\right)^x}{\left(\frac{1}{3}\right)^5} = 243 \cdot \left(\frac{1}{3}\right)^x.$$ For $x > 4$, the $y$-values of $h$ are $$h(x) = 243 \cdot \left(\frac{1}{3}\right)^x.$$ **Final answers:** - $t = -1$ - $A = (1, 0)$ - $f^{-1}(x) = \left(\frac{1}{3}\right)^x$ - Asymptote of $f^{-1}$: $y=0$ - $h(x) = 243 \cdot \left(\frac{1}{3}\right)^x$ for $x > 4$