To find the vertex of the quadratic function \( f(x) = -3x^2 + 6x - 2 \) using the completing-the-square method, we follow these steps:
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Factor out the coefficient of \( x^2 \) from the first two terms: \[ f(x) = -3(x^2 - 2x) - 2 \]
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Complete the square inside the parentheses. To do this, take half of the coefficient of \( x \) (which is \(-2\)), square it, and add and subtract that value inside the parentheses. Half of \(-2\) is \(-1\), and squaring it gives us \(1\): \[ f(x) = -3(x^2 - 2x + 1 - 1) - 2 \] This simplifies to: \[ f(x) = -3((x - 1)^2 - 1) - 2 \]
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Distribute the \(-3\): \[ f(x) = -3(x - 1)^2 + 3 - 2 \] So, we have: \[ f(x) = -3(x - 1)^2 + 1 \]
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Identify the vertex. The vertex form of a quadratic function is given by \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. From the equation, we see that: \[ h = 1 \quad \text{and} \quad k = 1 \] Thus, the vertex is \((1, 1)\).
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Determine whether it's a maximum or minimum. Since the coefficient of the squared term \(-3\) is negative, the parabola opens downward, indicating that the vertex represents a maximum point.
Therefore, the vertex of the function \( f(x) = -3x^2 + 6x - 2 \) is a maximum at the point \((1, 1)\).
The final answer is: Maximum at (1, 1).
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