1. **State the problem:** Given the function $h(x) = \sqrt{x - 1} + 2$, we need to find its domain and range, identify the end node, find the inverse function $h^{-1}(x)$ along with its domain and range, and plot the function. 2. **Find the domain of $h(x)$:** The expression inside the square root must be non-negative: $$x - 1 \geq 0$$ $$x \geq 1$$ So, the domain is $[1, \infty)$. 3. **Find the range of $h(x)$:** Since $\sqrt{x - 1} \geq 0$, the smallest value of $h(x)$ is when $x=1$: $$h(1) = \sqrt{1 - 1} + 2 = 0 + 2 = 2$$ As $x \to \infty$, $h(x) \to \infty$. So the range is $[2, \infty)$. 4. **Identify the end node:** The end node is the starting point of the function on the graph, which is at $(1, 2)$. 5. **Find the inverse function $h^{-1}(x)$:** Start with $$y = \sqrt{x - 1} + 2$$ Swap $x$ and $y$: $$x = \sqrt{y - 1} + 2$$ Isolate the square root: $$x - 2 = \sqrt{y - 1}$$ Square both sides: $$ (x - 2)^2 = y - 1 $$ Add 1 to both sides: $$ y = (x - 2)^2 + 1 $$ So, $$h^{-1}(x) = (x - 2)^2 + 1$$ 6. **Domain and range of $h^{-1}(x)$:** - Domain of $h^{-1}$ is the range of $h$, which is $[2, \infty)$. - Range of $h^{-1}$ is the domain of $h$, which is $[1, \infty)$. 7. **Complete the table:** Given $h(x)$ values 2, 3, 4, 5, find corresponding $x$ values: - For $h(x) = 2$: $$2 = \sqrt{x - 1} + 2 \Rightarrow \sqrt{x - 1} = 0 \Rightarrow x = 1$$ - For $h(x) = 3$: $$3 = \sqrt{x - 1} + 2 \Rightarrow \sqrt{x - 1} = 1 \Rightarrow x = 2$$ - For $h(x) = 4$: $$4 = \sqrt{x - 1} + 2 \Rightarrow \sqrt{x - 1} = 2 \Rightarrow x = 5$$ - For $h(x) = 5$: $$5 = \sqrt{x - 1} + 2 \Rightarrow \sqrt{x - 1} = 3 \Rightarrow x = 10$$ The missing negative $x$ value in the table is not valid since domain starts at 1. 8. **Plot $h(x)$:** The function starts at $(1, 2)$ and increases slowly as $x$ increases.