Let's analyze the function \( f(x) = (x + 1)^2 + 2 \) step-by-step to determine which characteristics apply:
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Domain:
- The expression \( (x + 1)^2 \) is defined for all real numbers \( x \). Therefore, the domain of \( f(x) \) is all real numbers.
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Range:
- The term \( (x + 1)^2 \) reaches a minimum value of 0 (when \( x = -1 \)). Thus, \( f(x) \) achieves a minimum value of \( 0 + 2 = 2 \). Therefore, the range is all real numbers greater than or equal to 2, not 1.
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Y-intercept:
- To find the y-intercept, we evaluate \( f(0) \): \[ f(0) = (0 + 1)^2 + 2 = 1^2 + 2 = 1 + 2 = 3. \]
- So the y-intercept is indeed 3.
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Transformation from the graph of \( y = x^2 \):
- The function \( f(x) = (x + 1)^2 + 2 \) represents a transformation of \( y = x^2 \):
- The graph shifts 1 unit to the left (due to \( x + 1 \)) and 2 units up (due to \( +2 \)).
- This statement is correct.
- The function \( f(x) = (x + 1)^2 + 2 \) represents a transformation of \( y = x^2 \):
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X-intercepts:
- To find the x-intercepts, we set \( f(x) = 0 \): \[ (x + 1)^2 + 2 = 0. \] However, since \( (x + 1)^2 \geq 0 \), \( (x + 1)^2 + 2 \geq 2 \), which means there are no x-intercepts.
- Therefore, the graph does not have two x-intercepts.
Based on this analysis, the characteristics that apply to the function \( f(x) = (x + 1)^2 + 2 \) are:
- The domain is all real numbers.
- The y-intercept is 3.
- The graph of the function is 1 unit up and 2 units to the left from the graph of \( y = x^2 \).
Thus, the correct answers are:
- The domain is all real numbers.
- The y-intercept is 3.
- The graph of the function is 1 unit up and 2 units to the left from the graph of \( y = x^2 \).
1 answer