1. **State the problem:** We need to analyze the function $$y=4x^2-81$$. 2. **Formula and rules:** This is a quadratic function in the form $$y=ax^2+bx+c$$ where $$a=4$$, $$b=0$$, and $$c=-81$$. 3. **Find the vertex:** The vertex of a parabola $$y=ax^2+bx+c$$ is at $$x=-\frac{b}{2a}$$. Since $$b=0$$, $$x=0$$. Calculate $$y$$ at $$x=0$$: $$y=4(0)^2-81=-81$$. So the vertex is at $$(0,-81)$$. 4. **Find the x-intercepts:** Set $$y=0$$: $$0=4x^2-81$$ $$4x^2=81$$ $$x^2=\frac{81}{4}$$ $$x=\pm\sqrt{\frac{81}{4}}=\pm\frac{9}{2}$$. 5. **Find the y-intercept:** Set $$x=0$$: $$y=4(0)^2-81=-81$$. 6. **Summary:** - Vertex at $$(0,-81)$$ - X-intercepts at $$x=\frac{9}{2}$$ and $$x=-\frac{9}{2}$$ - Y-intercept at $$(0,-81)$$ The parabola opens upwards because $$a=4>0$$. **Final answer:** The function $$y=4x^2-81$$ has vertex $$(0,-81)$$, x-intercepts at $$\pm\frac{9}{2}$$, and y-intercept at $$(0,-81)$$.