What is the minimum value of f(x) = 4x^2 - 16x + 15? (2022)

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Question: What is the minimum value of f(x) = 4x^2 - 16x + 15? (2022)

Options:

Correct Answer: 2

Exam Year: 2022

Solution:

The minimum occurs at x = -b/(2a) = 16/(2*4) = 2. f(2) = 4(2^2) - 16(2) + 15 = 1.

What is the minimum value of f(x) = 4x^2 - 16x + 15? (2022)

What is the minimum value of f(x) = 4x^2 - 16x + 15? (2022)

Step 1: Identify the function we need to analyze, which is f(x) = 4x^2 - 16x + 15.
Step 2: Recognize that this is a quadratic function in the form f(x) = ax^2 + bx + c, where a = 4, b = -16, and c = 15.
Step 3: To find the minimum value of a quadratic function, use the formula for the x-coordinate of the vertex: x = -b/(2a).
Step 4: Substitute the values of a and b into the formula: x = -(-16)/(2*4).
Step 5: Simplify the expression: x = 16/(8) = 2.
Step 6: Now, substitute x = 2 back into the function to find the minimum value: f(2) = 4(2^2) - 16(2) + 15.
Step 7: Calculate f(2): f(2) = 4(4) - 32 + 15 = 16 - 32 + 15.
Step 8: Simplify the calculation: f(2) = 16 - 32 + 15 = -16 + 15 = -1.
Step 9: Therefore, the minimum value of f(x) is -1, which occurs at x = 2.

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