Question

y=-x²+3=

Answer 1

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:

1. **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 \( (0, 3) \).

2. **Axis of Symmetry**: The axis of symmetry is the line \( x = 0 \).

3. **Y-intercept**: The y-intercept occurs where \( x = 0 \). Plugging in 0 gives \( y = 3 \), so the y-intercept is at (0, 3).

4. **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!