Sqrt Functions 0B7884

1. **Problem:** Sketch the graph of $$f(x) = \sqrt{x - 1} + 2$$ and find its domain and range. **Step 1:** The base function is $$y = \sqrt{x}$$. **Step 2:** The function is shifted right by 1 and up by 2. **Step 3:** The starting point (vertex) is at $$x = 1$$, so $$f(1) = \sqrt{1 - 1} + 2 = 0 + 2 = 2$$. **Step 4:** Domain: $$x - 1 \geq 0 \Rightarrow x \geq 1$$. **Step 5:** Range: Since $$\sqrt{x - 1} \geq 0$$, range is $$[2, \infty)$$. 2. **Problem:** Sketch the graph of $$f(x) = \sqrt{x + 3} - 4$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Shift left by 3 and down by 4. **Step 3:** Starting point at $$x = -3$$, $$f(-3) = \sqrt{-3 + 3} - 4 = 0 - 4 = -4$$. **Step 4:** Domain: $$x + 3 \geq 0 \Rightarrow x \geq -3$$. **Step 5:** Range: $$[-4, \infty)$$. 3. **Problem:** Sketch the graph of $$f(x) = -\sqrt{x - 2} + 1$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Shift right by 2, up by 1, and reflect over x-axis. **Step 3:** Starting point at $$x = 2$$, $$f(2) = -\sqrt{0} + 1 = 1$$. **Step 4:** Domain: $$x - 2 \geq 0 \Rightarrow x \geq 2$$. **Step 5:** Range: Since $$-\sqrt{x - 2} \leq 0$$, range is $$(-\infty, 1]$$. 4. **Problem:** Sketch the graph of $$f(x) = \sqrt{4 - x}$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Reflect inside the root and shift. **Step 3:** Domain: $$4 - x \geq 0 \Rightarrow x \leq 4$$. **Step 4:** Starting point at $$x = 4$$, $$f(4) = \sqrt{0} = 0$$. **Step 5:** Range: $$[0, \infty)$$. 5. **Problem:** Sketch the graph of $$f(x) = -\sqrt{5 - x} - 2$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Reflect inside root, reflect outside root, shift down by 2. **Step 3:** Domain: $$5 - x \geq 0 \Rightarrow x \leq 5$$. **Step 4:** Starting point at $$x = 5$$, $$f(5) = -\sqrt{0} - 2 = -2$$. **Step 5:** Range: Since $$-\sqrt{5 - x} \leq 0$$, range is $$(-\infty, -2]$$. 6. **Problem:** Sketch the graph of $$f(x) = \sqrt{x + 2}$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Shift left by 2. **Step 3:** Domain: $$x + 2 \geq 0 \Rightarrow x \geq -2$$. **Step 4:** Starting point at $$x = -2$$, $$f(-2) = 0$$. **Step 5:** Range: $$[0, \infty)$$. 7. **Problem:** Sketch the graph of $$f(x) = -\sqrt{x + 4} - 3$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Shift left by 4, reflect over x-axis, shift down by 3. **Step 3:** Domain: $$x + 4 \geq 0 \Rightarrow x \geq -4$$. **Step 4:** Starting point at $$x = -4$$, $$f(-4) = -0 - 3 = -3$$. **Step 5:** Range: $$(-\infty, -3]$$. 8. **Problem:** Sketch the graph of $$f(x) = \sqrt{1 - x} + 5$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Reflect inside root, shift up by 5. **Step 3:** Domain: $$1 - x \geq 0 \Rightarrow x \leq 1$$. **Step 4:** Starting point at $$x = 1$$, $$f(1) = 0 + 5 = 5$$. **Step 5:** Range: $$[5, \infty)$$. 9. **Problem:** Sketch the graph of $$f(x) = -\sqrt{x} + 6$$ and find its domain and range. **Step 1:** Base function $$y = \sqrt{x}$$. **Step 2:** Reflect over x-axis, shift up by 6. **Step 3:** Domain: $$x \geq 0$$. **Step 4:** Starting point at $$x = 0$$, $$f(0) = 6$$. **Step 5:** Range: $$(-\infty, 6]$$. 10. **Problem:** Sketch the graph of $$f(x) = \sqrt{x^2 - 9}$$ and find its domain and range. **Step 1:** Inside root must be non-negative: $$x^2 - 9 \geq 0$$. **Step 2:** Solve inequality: $$x^2 \geq 9 \Rightarrow |x| \geq 3$$. **Step 3:** Domain: $$(-\infty, -3] \cup [3, \infty)$$. **Step 4:** Starting points at $$x = -3$$ and $$x = 3$$, $$f(\pm 3) = 0$$. **Step 5:** Range: $$[0, \infty)$$. **Summary:** | Function | Domain | Range | Starting Point | |---|---|---|---| | $$\sqrt{x - 1} + 2$$ | $$[1, \infty)$$ | $$[2, \infty)$$ | $$(1, 2)$$ | | $$\sqrt{x + 3} - 4$$ | $$[-3, \infty)$$ | $$[-4, \infty)$$ | $$(-3, -4)$$ | | $$-\sqrt{x - 2} + 1$$ | $$[2, \infty)$$ | $$(-\infty, 1]$$ | $$(2, 1)$$ | | $$\sqrt{4 - x}$$ | $$(-\infty, 4]$$ | $$[0, \infty)$$ | $$(4, 0)$$ | | $$-\sqrt{5 - x} - 2$$ | $$(-\infty, 5]$$ | $$(-\infty, -2]$$ | $$(5, -2)$$ | | $$\sqrt{x + 2}$$ | $$[-2, \infty)$$ | $$[0, \infty)$$ | $$(-2, 0)$$ | | $$-\sqrt{x + 4} - 3$$ | $$[-4, \infty)$$ | $$(-\infty, -3]$$ | $$(-4, -3)$$ | | $$\sqrt{1 - x} + 5$$ | $$(-\infty, 1]$$ | $$[5, \infty)$$ | $$(1, 5)$$ | | $$-\sqrt{x} + 6$$ | $$[0, \infty)$$ | $$(-\infty, 6]$$ | $$(0, 6)$$ | | $$\sqrt{x^2 - 9}$$ | $$(-\infty, -3] \cup [3, \infty)$$ | $$[0, \infty)$$ | $$(-3, 0), (3, 0)$$ |