Determine the minimum value of f(x) = x^2 - 4x + 5. (2021)

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Question: Determine the minimum value of f(x) = x^2 - 4x + 5. (2021)

Options:

Correct Answer: 1

Exam Year: 2021

Solution:

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 5 = 1.

Determine the minimum value of f(x) = x^2 - 4x + 5. (2021)

Determine the minimum value of f(x) = x^2 - 4x + 5. (2021)

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.
Step 3: Find the vertex of the quadratic function, which gives the minimum value when a > 0.
Step 4: Use the formula for the x-coordinate of the vertex: x = -b/(2a). Here, a = 1 and b = -4.
Step 5: Calculate the x-coordinate of the vertex: 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): f(2) = 4 - 8 + 5 = 1.
Step 8: Conclude that the minimum value of f(x) is 1.

Quadratic Functions – Understanding the properties of quadratic functions, including how to find their minimum or maximum values using vertex form.
Vertex of a Parabola – Identifying the vertex of a parabola represented by a quadratic function, which indicates the minimum or maximum point.
Completing the Square – Using the method of completing the square to convert a quadratic function into vertex form.