Find the local minima of f(x) = x^2 - 4x + 5.

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Question: Find the local minima of f(x) = x^2 - 4x + 5.

Options:

Correct Answer: (2, 1)

Solution:

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, so local minima is (2, 1).

Find the local minima of f(x) = x^2 - 4x + 5.

Find the local minima of f(x) = x^2 - 4x + 5.

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 form of ax^2 + bx + c.
Step 3: Determine the coefficients: a = 1, b = -4, and c = 5.
Step 4: Use the formula for the x-coordinate of the vertex, which is x = -b/(2a).
Step 5: Substitute the values of a and b into the formula: x = -(-4)/(2*1) = 4/2 = 2.
Step 6: Now that we have the x-coordinate of the vertex (x = 2), we need to find the y-coordinate by substituting x back into the function.
Step 7: Calculate f(2) = 2^2 - 4*2 + 5 = 4 - 8 + 5 = 1.
Step 8: The vertex of the function is at the point (2, 1), which is the local minimum.

Quadratic Functions – Understanding the properties of quadratic functions, including how to find their vertex and determine local minima or maxima.
Vertex Formula – Using the vertex formula x = -b/(2a) to find the x-coordinate of the vertex for a quadratic function.
Function Evaluation – Evaluating the function at the vertex to find the corresponding y-coordinate.