1. **State the problem:** We are given the function $f(x) = \sqrt{x} - 6$ and want to understand its behavior and graph it. 2. **Recall the domain of the square root function:** The square root function $\sqrt{x}$ is defined only for $x \geq 0$ because we cannot take the square root of a negative number in the real number system. 3. **Analyze the function:** The function subtracts 6 from $\sqrt{x}$, so it shifts the graph of $\sqrt{x}$ down by 6 units. 4. **Domain:** Since $\sqrt{x}$ requires $x \geq 0$, the domain of $f(x)$ is $[0, \infty)$. 5. **Range:** The smallest value of $\sqrt{x}$ is 0 (at $x=0$), so the smallest value of $f(x)$ is $0 - 6 = -6$. As $x$ increases, $\sqrt{x}$ increases without bound, so the range of $f(x)$ is $[-6, \infty)$. 6. **Intercepts:** - **x-intercept:** Set $f(x) = 0$: $$\sqrt{x} - 6 = 0 \implies \sqrt{x} = 6 \implies x = 36.$$ So the x-intercept is at $(36, 0)$. - **y-intercept:** Since the domain starts at 0, the y-intercept is at $f(0) = \sqrt{0} - 6 = -6$, so the point is $(0, -6)$. 7. **Graph features:** The graph starts at $(0, -6)$ and increases slowly, curving upward as $x$ increases. 8. **Summary:** The function $f(x) = \sqrt{x} - 6$ is defined for $x \geq 0$, has a y-intercept at $(0, -6)$, an x-intercept at $(36, 0)$, and its graph is the square root curve shifted down by 6 units. **Final answer:** The function $f(x) = \sqrt{x} - 6$ has domain $[0, \infty)$, range $[-6, \infty)$, x-intercept at $(36, 0)$, and y-intercept at $(0, -6)$.