1. **State the problem:** Find the vertex, y-intercept, and x-intercepts of the parabola given by the equation $$y = x^2 - 4$$. 2. **Vertex:** The vertex form of a parabola is $$y = a(x-h)^2 + k$$ where $$(h,k)$$ is the vertex. For $$y = x^2 - 4$$, it is already in vertex form with $$a=1$$, $$h=0$$, and $$k=-4$$. So, the vertex is at $$(0, -4)$$. 3. **Y-intercept:** The y-intercept occurs when $$x=0$$. Substitute $$x=0$$ into the equation: $$y = 0^2 - 4 = -4$$ So, the y-intercept is $$(0, -4)$$. 4. **X-intercepts:** The x-intercepts occur when $$y=0$$. Set $$y=0$$ and solve for $$x$$: $$0 = x^2 - 4$$ Add 4 to both sides: $$4 = x^2$$ Take the square root of both sides: $$x = \pm \sqrt{4}$$ $$x = \pm 2$$ So, the x-intercepts are $$(-2, 0)$$ and $$(2, 0)$$. **Final answers:** - Vertex: $$(0, -4)$$ - Y-intercept: $$(0, -4)$$ - X-intercepts: $$(-2, 0)$$ and $$(2, 0)$$