Find the minimum value of f(x) = x^2 - 4x + 6. (2021)

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Question: Find the minimum value of f(x) = x^2 - 4x + 6. (2021)

Options:

Correct Answer: 2

Exam Year: 2021

Solution:

The vertex form gives the minimum at x = 2. f(2) = 2.

Find the minimum value of f(x) = x^2 - 4x + 6. (2021)

Find the minimum value of f(x) = x^2 - 4x + 6. (2021)

Step 1: Identify the function you need to analyze, which is f(x) = x^2 - 4x + 6.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c.
Step 3: Determine the coefficients: a = 1, b = -4, and c = 6.
Step 4: Use the formula for the x-coordinate of the vertex, which is x = -b/(2a).
Step 5: Substitute the values of b and a into the formula: x = -(-4)/(2*1) = 4/2 = 2.
Step 6: Now that you have x = 2, substitute this value back into the function to find the minimum value: f(2) = (2)^2 - 4*(2) + 6.
Step 7: Calculate f(2): f(2) = 4 - 8 + 6 = 2.
Step 8: Conclude that the minimum value of f(x) is 2, which occurs at x = 2.

Quadratic Functions – Understanding the properties of quadratic functions, including how to find their minimum or maximum values using vertex form.
Vertex Form – Recognizing that the vertex of a parabola represented by a quadratic function gives the minimum or maximum value.