Answer:
To find the vertex of the quadratic function \( f(x) = -3x^2 + 6x - 2 \) using the completing-the-square method, follow these steps:
1. Write the quadratic expression in the form \( a(x - h)^2 + k \).
2. The vertex is given by the coordinates \((h, k)\), where \(h\) and \(k\) are the values obtained.
Starting with the given function:
\[ f(x) = -3x^2 + 6x - 2 \]
1. Factor out the coefficient of \(x^2\) from the \(x^2\) and \(x\) terms:
\[ f(x) = -3(x^2 - 2x) - 2 \]
2. Complete the square inside the parentheses:
\[ f(x) = -3(x^2 - 2x + 1 - 1) - 2 \]
\[ f(x) = -3(x - 1)^2 + 3 - 2 \]
\[ f(x) = -3(x - 1)^2 - 1 \]
Now, the vertex form is \( a(x - h)^2 + k \) where \((h, k)\) is the vertex. In this case, the vertex is \((1, -1)\).
So, the correct answer is:
C. Maximum at (–1, 2)
