Find the minimum value of f(x) = 4x^2 - 8x + 3. (2022)

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Question: Find the minimum value of f(x) = 4x^2 - 8x + 3. (2022)

Options:

Correct Answer: 1

Exam Year: 2022

Solution:

The vertex is at x = 8/(2*4) = 1. The minimum value is f(1) = 4(1)^2 - 8(1) + 3 = -1.

Find the minimum value of f(x) = 4x^2 - 8x + 3. (2022)

Find the minimum value of f(x) = 4x^2 - 8x + 3. (2022)

Step 1: Identify the function we need to analyze, which is f(x) = 4x^2 - 8x + 3.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c, where a = 4, b = -8, and c = 3.
Step 3: To find the vertex of the parabola (which gives the minimum value since a > 0), use the formula x = -b/(2a).
Step 4: Substitute the values of a and b into the formula: x = -(-8)/(2*4) = 8/(8) = 1.
Step 5: Now that we have x = 1, we need to find the minimum value of the function by substituting x back into f(x).
Step 6: Calculate f(1) = 4(1)^2 - 8(1) + 3.
Step 7: Simplify the expression: f(1) = 4(1) - 8 + 3 = 4 - 8 + 3 = -1.
Step 8: Therefore, the minimum value of f(x) is -1.

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