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Determine the minimum value of the function f(x) = x^2 - 4x + 5.
Determine the minimum value of the function f(x) = x^2 - 4x + 5.
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Practice Questions
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Q1
Determine the minimum value of the function f(x) = x^2 - 4x + 5.
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The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1. Thus, the minimum value is 1.
Questions & Step-by-step Solutions
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Q
Q: Determine the minimum value of the function f(x) = x^2 - 4x + 5.
Solution:
The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1. Thus, the minimum value is 1.
Steps: 7
Show Steps
Step 1: Identify the function we need to analyze, which is f(x) = x^2 - 4x + 5.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c, where a = 1, b = -4, and c = 5.
Step 3: Find the x-coordinate of the vertex using the formula x = -b/(2a). Here, b = -4 and a = 1.
Step 4: Calculate x = -(-4)/(2*1) = 4/2 = 2.
Step 5: Now, substitute x = 2 back into the function to find the minimum value: f(2) = 2^2 - 4*2 + 5.
Step 6: Calculate f(2) = 4 - 8 + 5 = 1.
Step 7: Conclude that the minimum value of the function f(x) is 1.
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