What is the minimum value of the function f(x) = 4x^2 - 16x + 20? (2021) 2021

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Question: What is the minimum value of the function f(x) = 4x^2 - 16x + 20? (2021) 2021

Options:

Correct Answer: 5

Exam Year: 2021

Solution:

The vertex occurs at x = 2. f(2) = 4(2^2) - 16(2) + 20 = 4.

What is the minimum value of the function f(x) = 4x^2 - 16x + 20? (2021) 2021

What is the minimum value of the function f(x) = 4x^2 - 16x + 20? (2021) 2021

Step 1: Identify the function we need to analyze, which is f(x) = 4x^2 - 16x + 20.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c, where a = 4, b = -16, and c = 20.
Step 3: Determine the x-coordinate of the vertex using the formula x = -b / (2a). Here, b = -16 and a = 4.
Step 4: Calculate x = -(-16) / (2 * 4) = 16 / 8 = 2.
Step 5: Now, substitute x = 2 back into the function to find the minimum value: f(2) = 4(2^2) - 16(2) + 20.
Step 6: Calculate f(2) = 4(4) - 32 + 20 = 16 - 32 + 20 = 4.
Step 7: Conclude that the minimum value of the function f(x) is 4.

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