To analyze the functions \( f(x) = \sqrt{x + 1} - 2 \) and \( g(x) = \sqrt{x + 1} + 3 - 2 \), we first simplify \( g(x) \):
\[ g(x) = \sqrt{x + 1} + 1 \]
Now let's find the features of each graph.
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Domain:
- The domain of \( f(x) = \sqrt{x + 1} - 2 \) is \( x + 1 \geq 0 \) which means \( x \geq -1 \). The domain is \( [-1, \infty) \).
- The domain of \( g(x) = \sqrt{x + 1} + 1 \) is also \( x + 1 \geq 0 \), which gives the same domain \( [-1, \infty) \).
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Y-Intercept:
- The y-intercept for \( f(x) \) is found by evaluating \( f(0) \): \[ f(0) = \sqrt{0 + 1} - 2 = 1 - 2 = -1 \]
- The y-intercept for \( g(x) \) is found by evaluating \( g(0) \): \[ g(0) = \sqrt{0 + 1} + 1 = 1 + 1 = 2 \]
- Thus, the y-intercepts are different: \( f(x) \) at \(-1\) and \( g(x) \) at \(2\).
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X-Intercept:
- The x-intercept for \( f(x) \) occurs when \( f(x) = 0 \): \[ \sqrt{x + 1} - 2 = 0 \Rightarrow \sqrt{x + 1} = 2 \Rightarrow x + 1 = 4 \Rightarrow x = 3 \]
- The x-intercept for \( g(x) \): \[ g(x) = 0 \Rightarrow \sqrt{x + 1} + 1 = 0 \Rightarrow \sqrt{x + 1} = -1 \] This has no solutions since a square root cannot be negative.
- Thus, the x-intercepts are different: \( f(x) \) has an x-intercept at \( x = 3 \) and \( g(x) \) has none.
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Range:
- The range of \( f(x) \): The function starts at \( -1 \) when \( x = -1 \) and approaches infinity as \( x \) increases, so its range is \( [-1, \infty) \).
- The range of \( g(x) \): The function starts at \( 2 \) when \( x = -1 \) and also approaches infinity, so its range is \( [2, \infty) \).
- Thus, the ranges are different.
Comparing all features:
- The graphs have the same domain.
- Different y-intercepts.
- Different x-intercepts.
- Different ranges.
Conclusion: The statement "The graphs have the same domains" is true.