Determine the minimum value of the function f(x) = x^2 - 4x + 6. (2020)

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Question: Determine the minimum value of the function f(x) = x^2 - 4x + 6. (2020)

Options:

Correct Answer: 2

Exam Year: 2020

Solution:

The function is a upward-opening parabola. The minimum occurs at x = -b/(2a) = 4/(2*1) = 2. f(2) = 2^2 - 4(2) + 6 = 2.

Determine the minimum value of the function f(x) = x^2 - 4x + 6. (2020)

Determine the minimum value of the function f(x) = x^2 - 4x + 6. (2020)

Step 1: Identify the function we are working with, which is f(x) = x^2 - 4x + 6.
Step 2: Recognize that this function is a quadratic function, which means it forms a parabola when graphed.
Step 3: Determine the direction of the parabola. Since the coefficient of x^2 is positive (1), the parabola opens upwards.
Step 4: Find the x-coordinate of the vertex of the parabola, which gives the minimum value. Use the formula x = -b/(2a). Here, a = 1 and b = -4.
Step 5: Substitute the values into the formula: x = -(-4)/(2*1) = 4/2 = 2.
Step 6: Now, calculate the minimum value of the function by substituting x = 2 back into the function: f(2) = 2^2 - 4(2) + 6.
Step 7: Simplify the expression: f(2) = 4 - 8 + 6 = 2.
Step 8: Conclude that the minimum value of the function f(x) is 2.

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 parabola.
Function Evaluation – Evaluating the function at the vertex to find the minimum value.