For the function f(x) = x^2 - 6x + 8, find the x-coordinate of the vertex.

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Question: For the function f(x) = x^2 - 6x + 8, find the x-coordinate of the vertex.

Options:

Correct Answer: 3

Solution:

The x-coordinate of the vertex is given by x = -b/(2a) = 6/(2*1) = 3.

For the function f(x) = x^2 - 6x + 8, find the x-coordinate of the vertex.

For the function f(x) = x^2 - 6x + 8, find the x-coordinate of the vertex.

Step 1: Identify the coefficients a, b, and c from the function f(x) = x^2 - 6x + 8. Here, a = 1, b = -6, and c = 8.
Step 2: Use the formula for the x-coordinate of the vertex, which is x = -b/(2a).
Step 3: Substitute the value of b into the formula. Since b = -6, we have x = -(-6)/(2*1).
Step 4: Simplify the expression. This becomes x = 6/(2*1).
Step 5: Calculate the denominator. 2*1 = 2.
Step 6: Now, divide 6 by 2. This gives x = 3.
Step 7: Therefore, the x-coordinate of the vertex is 3.

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