What is the minimum value of the function f(x) = x^2 - 4x + 5?
Practice Questions
1 question
Q1
What is the minimum value of the function f(x) = x^2 - 4x + 5?
1
0
2
4
The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 5 = 1.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the minimum value of the function f(x) = x^2 - 4x + 5?
Solution: The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 5 = 1.
Steps: 7
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.
