Find the maximum value of f(x) = -x^2 + 4x + 5. (2021)

₹0.0

Question: Find the maximum value of f(x) = -x^2 + 4x + 5. (2021)

Options:

Correct Answer: 7

Exam Year: 2021

Solution:

The vertex is at x = -4/(2*(-1)) = 2. The maximum value is f(2) = -2^2 + 4*2 + 5 = 7.

Find the maximum value of f(x) = -x^2 + 4x + 5. (2021)

Find the maximum value of f(x) = -x^2 + 4x + 5. (2021)

Step 1: Identify the function we need to analyze, which is f(x) = -x^2 + 4x + 5.
Step 2: Recognize that this is a quadratic function in the form of ax^2 + bx + c, where a = -1, b = 4, and c = 5.
Step 3: Find the x-coordinate of the vertex using the formula x = -b/(2a). Here, b = 4 and a = -1.
Step 4: Calculate the x-coordinate: x = -4/(2*(-1)) = -4/-2 = 2.
Step 5: Now, substitute x = 2 back into the function to find the maximum value: f(2) = -2^2 + 4*2 + 5.
Step 6: Calculate f(2): f(2) = -4 + 8 + 5 = 4 + 5 = 9.
Step 7: Therefore, the maximum value of the function f(x) is 9.

Quadratic Functions – Understanding the properties of quadratic functions, including finding the vertex and determining maximum or minimum 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 maximum or minimum value.