To compare the functions \( f(x) = \sqrt{x + 1} - 2 \) and \( g(x) = \sqrt{x + 1} + 3 - 2 \), we can analyze their features: x-intercepts, domains, y-intercepts, and ranges.
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X-intercepts:
- To find the x-intercepts, set \( f(x) = 0 \) and \( g(x) = 0 \).
- For \( f(x) = \sqrt{x + 1} - 2 = 0 \): \[ \sqrt{x + 1} = 2 \implies x + 1 = 4 \implies x = 3. \]
- For \( g(x) = \sqrt{x + 1} + 1 = 0 \): \[ \sqrt{x + 1} = -1 \text{ (not possible, since the square root cannot be negative)}. \]
- Thus, the x-intercept of \( g(x) \) does not exist and is different from that of \( f(x) \).
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Domains:
- Both functions contain \( \sqrt{x + 1} \), which requires \( x + 1 \geq 0 \) or \( x \geq -1 \).
- So the domain for both functions is \( x \geq -1 \).
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Y-intercepts:
- To find the y-intercepts, evaluate \( f(0) \) and \( g(0) \):
- For \( f(0) \): \[ f(0) = \sqrt{0 + 1} - 2 = 1 - 2 = -1. \]
- For \( g(0) \): \[ g(0) = \sqrt{0 + 1} + 3 - 2 = 1 + 1 = 2. \]
- Thus, the y-intercepts are \( (-1) \) for \( f \) and \( (2) \) for \( g \), so they are different.
- To find the y-intercepts, evaluate \( f(0) \) and \( g(0) \):
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Ranges:
- The range of \( f(x) \): \[ f(x) \text{ starts at } -1 \text{ and increases, approaching infinity. Thus, its range is } [-1, \infty). \]
- The range of \( g(x) \): \[ g(x) \text{ starts at } 2 \text{ and also increases, approaching infinity. Thus, its range is } [2, \infty). \]
- Since the lowest values in their ranges are different, this means their ranges are also different.
Conclusion:
The only feature that is true about the graphs is that they have the same domains. Therefore, the correct answer is:
The graphs have the same domains.