Find the minimum value of the function f(x) = x^2 - 4x + 5.
Practice Questions
1 question
Q1
Find the minimum value of the function f(x) = x^2 - 4x + 5.
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The vertex of the parabola occurs at x = 2. f(2) = 2^2 - 4(2) + 5 = 1, which is the minimum value.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the minimum value of the function f(x) = x^2 - 4x + 5.
Solution: The vertex of the parabola occurs at x = 2. f(2) = 2^2 - 4(2) + 5 = 1, which is the minimum value.
Steps: 7
Step 1: Identify the function you need to analyze, which is f(x) = x^2 - 4x + 5.
Step 2: Recognize that this function is a quadratic function, which means it forms a parabola.
Step 3: Determine the formula to find the x-coordinate of the vertex of the parabola, which is x = -b/(2a). Here, a = 1 and b = -4.
Step 4: Plug in the values of a and b into the formula: x = -(-4)/(2*1) = 4/2 = 2.
Step 5: Now that you have the x-coordinate of the vertex (x = 2), substitute this value back into the function to find the minimum value: f(2) = 2^2 - 4(2) + 5.
Step 6: Calculate f(2): f(2) = 4 - 8 + 5 = 1.
Step 7: Conclude that the minimum value of the function f(x) is 1.
