1. **State the problem:** We need to graph the function $$y=\frac{7}{8}x^3$$ and plot five points: one at $$x=0$$, two with negative $$x$$ values, and two with positive $$x$$ values. 2. **Formula and rules:** The function is a cubic function of the form $$y=ax^3$$ where $$a=\frac{7}{8}$$. Cubic functions are odd functions and symmetric about the origin. 3. **Calculate points:** - For $$x=0$$: $$y=\frac{7}{8}(0)^3=0$$, point is $$(0,0)$$. - For $$x=-2$$: $$y=\frac{7}{8}(-2)^3=\frac{7}{8}(-8)=-7$$, point is $$(-2,-7)$$. - For $$x=-1$$: $$y=\frac{7}{8}(-1)^3=\frac{7}{8}(-1)=-\frac{7}{8}$$, point is $$(-1,-\frac{7}{8})$$. - For $$x=1$$: $$y=\frac{7}{8}(1)^3=\frac{7}{8}$$, point is $$(1,\frac{7}{8})$$. - For $$x=2$$: $$y=\frac{7}{8}(2)^3=\frac{7}{8}(8)=7$$, point is $$(2,7)$$. 4. **Summary:** The five points to plot are $$(0,0), (-2,-7), (-1,-\frac{7}{8}), (1,\frac{7}{8}), (2,7)$$. 5. **Graph behavior:** The cubic function increases steeply for large positive and negative $$x$$ values, passing through the origin. Final answer: The function $$y=\frac{7}{8}x^3$$ with points $$(0,0), (-2,-7), (-1,-\frac{7}{8}), (1,\frac{7}{8}), (2,7)$$ plotted on the graph.