1. **State the problem:** We are given the function $h(x) = 1 - \sqrt{x}$ and we want to understand its behavior. 2. **Recall the square root function:** The square root function $\sqrt{x}$ is defined for $x \geq 0$ and outputs non-negative values. 3. **Analyze the function:** Since $h(x) = 1 - \sqrt{x}$, as $x$ increases, $\sqrt{x}$ increases, so $h(x)$ decreases. 4. **Domain:** The domain of $h(x)$ is $x \geq 0$ because $\sqrt{x}$ is only defined for non-negative $x$. 5. **Range:** Since $\sqrt{x} \geq 0$, the maximum value of $h(x)$ is when $x=0$, giving $h(0) = 1 - 0 = 1$. As $x \to \infty$, $\sqrt{x} \to \infty$, so $h(x) \to -\infty$. Thus, the range is $(-\infty, 1]$. 6. **Intercepts:** - **y-intercept:** At $x=0$, $h(0) = 1$. - **x-intercept:** Solve $h(x) = 0$: $$ 0 = 1 - \sqrt{x} \\ \sqrt{x} = 1 \\ x = 1^2 = 1 $$ So the x-intercept is at $x=1$. 7. **Summary:** The function starts at $(0,1)$ and decreases as $x$ increases, crossing the x-axis at $(1,0)$ and continuing downward.