What is the maximum value of the function f(x) = -x^2 + 4x + 1?
Practice Questions
1 question
Q1
What is the maximum value of the function f(x) = -x^2 + 4x + 1?
5
6
7
8
The vertex occurs at x = -b/(2a) = 4/(-2) = 2. f(2) = -2^2 + 4*2 + 1 = 6.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the maximum value of the function f(x) = -x^2 + 4x + 1?
Solution: The vertex occurs at x = -b/(2a) = 4/(-2) = 2. f(2) = -2^2 + 4*2 + 1 = 6.
Steps: 8
Step 1: Identify the function we are working with, which is f(x) = -x^2 + 4x + 1.
Step 2: Recognize that this is a quadratic function in the form f(x) = ax^2 + bx + c, where a = -1, b = 4, and c = 1.
Step 3: To find the maximum 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).
Step 5: Calculate the value: x = -4 / -2 = 2. This means the vertex occurs at x = 2.
Step 6: Now, we need to find the maximum value of the function at this x-coordinate. Substitute x = 2 back into the function: f(2) = -2^2 + 4*2 + 1.
Step 7: Calculate f(2): f(2) = -4 + 8 + 1 = 5.
Step 8: Therefore, the maximum value of the function f(x) is 5.
