What is the minimum value of f(x) = x^2 - 4x + 6? (2022)

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Question: What is the minimum value of f(x) = x^2 - 4x + 6? (2022)

Options:

Correct Answer: 3

Solution:

The vertex occurs at x = 2. f(2) = 2.

What is the minimum value of f(x) = x^2 - 4x + 6? (2022)

What is the minimum value of f(x) = x^2 - 4x + 6? (2022)

Step 1: Identify the function we are working with, which is f(x) = x^2 - 4x + 6.
Step 2: Recognize that this is a quadratic function in the form of f(x) = ax^2 + bx + c, where a = 1, b = -4, and c = 6.
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) + 6.
Step 6: Calculate f(2) = 4 - 8 + 6 = 2.
Step 7: Conclude that the minimum value of f(x) is 2.

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