What is the maximum value of the quadratic function f(x) = -x^2 + 6x - 8?

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Question: What is the maximum value of the quadratic function f(x) = -x^2 + 6x - 8?

Options:

Correct Answer: 4

Solution:

The vertex form gives the maximum value at x = 3, f(3) = -3^2 + 6*3 - 8 = 4.

What is the maximum value of the quadratic function f(x) = -x^2 + 6x - 8?

What is the maximum value of the quadratic function f(x) = -x^2 + 6x - 8?

Step 1: Identify the quadratic function, which is f(x) = -x^2 + 6x - 8.
Step 2: Recognize that this is a quadratic function in the standard form ax^2 + bx + c, where a = -1, b = 6, and c = -8.
Step 3: Since the coefficient of x^2 (a) is negative, the parabola opens downwards, meaning it has a maximum value.
Step 4: To find the x-coordinate of the vertex (maximum point), use the formula x = -b/(2a).
Step 5: Substitute the values of a and b: x = -6/(2 * -1) = 3.
Step 6: Now, substitute x = 3 back into the function to find the maximum value: f(3) = -3^2 + 6*3 - 8.
Step 7: Calculate f(3): f(3) = -9 + 18 - 8 = 1.
Step 8: Therefore, the maximum value of the function is 1.

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