1. **State the problem:** We need to graph the radical functions and identify their domain and range. 2. **Function a: $f(x) = \sqrt{x} - 5$** - The square root function $\sqrt{x}$ is defined only for $x \geq 0$. - The domain of $f(x)$ is therefore $[0, \infty)$. - The range is shifted down by 5, so the range is $[-5, \infty)$. 3. **Function b: $h(x) = \sqrt{x - 2}$** - The expression under the square root must be non-negative: $x - 2 \geq 0$. - So the domain is $[2, \infty)$. - The range is $[0, \infty)$ because square root outputs non-negative values. 4. **Intermediate work for $f(x)$:** - Evaluate $f(x)$ at given $x$ values: - $f(0) = \sqrt{0} - 5 = 0 - 5 = -5$ - $f(1) = \sqrt{1} - 5 = 1 - 5 = -4$ - $f(2) = \sqrt{2} - 5 \approx 1.414 - 5 = -3.586$ - $f(4) = \sqrt{4} - 5 = 2 - 5 = -3$ - $f(8) = \sqrt{8} - 5 \approx 2.828 - 5 = -2.172$ - $f(9) = \sqrt{9} - 5 = 3 - 5 = -2$ 5. **Intermediate work for $h(x)$:** - Evaluate $h(x)$ at given $x$ values: - $h(2) = \sqrt{2 - 2} = \sqrt{0} = 0$ - $h(3) = \sqrt{3 - 2} = \sqrt{1} = 1$ - $h(6) = \sqrt{6 - 2} = \sqrt{4} = 2$ - $h(8) = \sqrt{8 - 2} = \sqrt{6} \approx 2.449$ - $h(11) = \sqrt{11 - 2} = \sqrt{9} = 3$ 6. **Summary:** - $f(x) = \sqrt{x} - 5$ has domain $[0, \infty)$ and range $[-5, \infty)$. - $h(x) = \sqrt{x - 2}$ has domain $[2, \infty)$ and range $[0, \infty)$. These functions are graphed as the square root curve shifted vertically or horizontally accordingly.