Question: User: y=-x^2+3
1. **State the problem:** We are given the quadratic function $$y = -x^2 + 3$$ and want to analyze its properties.
2. **Formula and rules:** This is a quadratic function in the form $$y = ax^2 + bx + c$$ where $$a = -1$$, $$b = 0$$, and $$c = 3$$.
- Since $$a < 0$$, the parabola opens downward.
- The vertex form of a parabola is $$y = a(x - h)^2 + k$$ where $$(h,k)$$ is the vertex.
3. **Find the vertex:**
- The vertex $$h$$ is given by $$h = -\frac{b}{2a} = -\frac{0}{2(-1)} = 0$$.
- Substitute $$h=0$$ into the function to find $$k$$:
$$k = -0^2 + 3 = 3$$.
- So the vertex is at $$(0, 3)$$.
4. **Find the y-intercept:**
- The y-intercept occurs when $$x=0$$:
$$y = -0^2 + 3 = 3$$.
5. **Find the x-intercepts:**
- Set $$y=0$$ and solve for $$x$$:
$$0 = -x^2 + 3$$
$$x^2 = 3$$
$$x = \pm \sqrt{3}$$.
6. **Summary:**
- The parabola opens downward.
- Vertex at $$(0, 3)$$.
- Y-intercept at $$(0, 3)$$.
- X-intercepts at $$x = \sqrt{3}$$ and $$x = -\sqrt{3}$$.
This completes the analysis of the quadratic function $$y = -x^2 + 3$$.