Parabola Graph Fec38D

1. **State the problem:** We need to graph the parabola given by the equation $$y=\frac{1}{4}x^2$$ and plot five points: the vertex, two points to the left, and two points to the right of the vertex. 2. **Identify the vertex:** The parabola is in the form $$y=ax^2$$ with $$a=\frac{1}{4}$$. The vertex of this parabola is at the origin $$ (0,0) $$. 3. **Calculate points to the left and right of the vertex:** - For $$x=-2$$: $$y=\frac{1}{4}(-2)^2=\frac{1}{4}\times 4=1$$, so the point is $$(-2,1)$$. - For $$x=-4$$: $$y=\frac{1}{4}(-4)^2=\frac{1}{4}\times 16=4$$, so the point is $$(-4,4)$$. - For $$x=2$$: $$y=\frac{1}{4}(2)^2=\frac{1}{4}\times 4=1$$, so the point is $$(2,1)$$. - For $$x=4$$: $$y=\frac{1}{4}(4)^2=\frac{1}{4}\times 16=4$$, so the point is $$(4,4)$$. 4. **Summary of points:** - Vertex: $$(0,0)$$ - Left points: $$(-2,1)$$ and $$(-4,4)$$ - Right points: $$(2,1)$$ and $$(4,4)$$ 5. **Graph shape:** The parabola opens upward because $$a=\frac{1}{4} > 0$$. The points plotted confirm the shape and symmetry about the y-axis. 6. **Final answer:** The parabola $$y=\frac{1}{4}x^2$$ with vertex at $$(0,0)$$ and points $$(\pm 2,1)$$, $$(\pm 4,4)$$ is graphed as an upward-opening curve symmetric about the y-axis.