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

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

Options:

Correct Answer: 1

Solution:

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1. Thus, the minimum value is 1.

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

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

Correct Answer: 1

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, where a = 1, b = -4, and c = 5.
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 + 5.
Step 6: Calculate f(2) = 4 - 8 + 5 = 1.
Step 7: Conclude that the minimum value of the function f(x) is 1.

Quadratic Functions – Understanding the properties of quadratic functions, including their vertex and minimum/maximum values.
Vertex Formula – Using the vertex formula x = -b/(2a) to find the x-coordinate of the vertex of a quadratic function.
Function Evaluation – Evaluating the function at the vertex to determine the minimum or maximum value.