It looks like you're presenting a quadratic function, but there seems to be a slight issue with the equation format. The equation can be recognized as:
\[ y = -x^2 + 3 \]
This is a downward-opening parabola with the following characteristics:
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Vertex: The vertex of this parabola can be found using the formula for the vertex of a quadratic function, which is \( (-b/2a, f(-b/2a)) \). In this case, \( a = -1 \) and \( b = 0 \), so the vertex is at \( (0, 3) \).
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Axis of Symmetry: The axis of symmetry is the line \( x = 0 \).
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Y-intercept: The y-intercept occurs where \( x = 0 \). Plugging in 0 gives \( y = 3 \), so the y-intercept is at (0, 3).
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X-intercepts: To find the x-intercepts, you set \( y = 0 \) and solve for \( x \): \[ 0 = -x^2 + 3 \Rightarrow x^2 = 3 \Rightarrow x = \pm \sqrt{3} \] So the x-intercepts are at \( (\sqrt{3}, 0) \) and \( (-\sqrt{3}, 0) \).
If you have any other specific questions or if you are looking for a certain transformation or property of this function, feel free to ask!