What is the minimum value of the function f(x) = x^2 - 4x + 5?

Practice Questions

Q1
What is the minimum value of the function f(x) = x^2 - 4x + 5?
  1. 1
  2. 0
  3. 2
  4. 4

Questions & Step-by-Step Solutions

What is the minimum value of the function f(x) = x^2 - 4x + 5?
Correct Answer: 1
  • 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 standard form ax^2 + bx + c, where a = 1, b = -4, and c = 5.
  • Step 3: To find the minimum value, we need to find the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b/(2a).
  • Step 4: Substitute the values of a and b into the formula: x = -(-4)/(2*1) = 4/2 = 2.
  • Step 5: Now that we have x = 2, we need to find the corresponding y-value (which is the minimum value of the function) by substituting x back into the function: f(2) = 2^2 - 4(2) + 5.
  • Step 6: Calculate f(2): f(2) = 4 - 8 + 5 = 1.
  • Step 7: Therefore, the minimum value of the function f(x) is 1.
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