What is the minimum value of f(x) = x^2 - 4x + 5? (2020)
Practice Questions
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What is the minimum value of f(x) = x^2 - 4x + 5? (2020)
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Questions & Step-by-Step Solutions
What is the minimum value of f(x) = x^2 - 4x + 5? (2020)
Step 1: Identify the function we are working with, which is f(x) = x^2 - 4x + 5.
Step 2: Recognize that this is a quadratic function in the standard form ax^2 + bx + c, where a = 1, b = -4, and c = 5.
Step 3: Understand that the graph of a quadratic function is a parabola. Since a (1) is positive, the parabola opens upwards, meaning it has a minimum point.
Step 4: To find the x-coordinate of the vertex (minimum point), use the formula x = -b/(2a). Here, b = -4 and a = 1.
Step 5: Calculate x = -(-4)/(2*1) = 4/2 = 2.
Step 6: Now, substitute x = 2 back into the function to find the minimum value: f(2) = (2)^2 - 4*(2) + 5.
Step 7: Calculate f(2) = 4 - 8 + 5 = 1.
Step 8: Therefore, the minimum value of f(x) is 1, which occurs at x = 2.
