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$y = 3(x - 4)^2 - 2$
$y = 3(x + 4)^2 - 2$
$y = -3(x - 4)^2 - 2$
$y = -3(x + 4)^2 - 2$
Step1: Analyze the parabola's direction
The parabola opens downward, so the coefficient of the squared term should be negative. This eliminates the options with a positive coefficient (\(y = 3(x - 4)^2 - 2\) and \(y = 3(x + 4)^2 - 2\)).
Step2: Find the vertex form
The vertex form of a parabola is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. From the graph, the vertex is at \((4, -2)\) (since it's shifted 4 units right and 2 units down). Also, the parabola opens downward, so \(a\) is negative. The equation \(y = -3(x - 4)^2 - 2\) has \(a=-3\) (negative), \(h = 4\), and \(k=-2\), which matches the vertex and the direction.
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\(y = -3(x - 4)^2 - 2\) (the second option in the lower - left group of equations)