1. **State the problem:** We are given the equation $$x^2 - 3y = 31$$ and asked to analyze it. 2. **Rewrite the equation to express y in terms of x:** $$x^2 - 3y = 31$$ Subtract $$x^2$$ from both sides: $$- 3y = 31 - x^2$$ Divide both sides by $$-3$$: $$y = \frac{x^2 - 31}{3}$$ 3. **Interpretation:** The equation is not a circle; it represents a parabola opening upwards because $$y$$ is quadratic in $$x$$. 4. **Graphing properties:** - Vertex occurs where $$x=0$$: $$y = \frac{0 - 31}{3} = -\frac{31}{3}$$ Vertex is at $$(0,-\frac{31}{3})$$. - The parabola opens upwards since the coefficient of $$x^2$$ in $$y$$ is positive ($$\frac{1}{3}$$). 5. **No circle present:** The claim of a circle centered top-right does not match this equation's form. **Final answer:** The equation $$x^2 - 3y = 31$$ can be rewritten as $$y = \frac{x^2 - 31}{3}$$, representing a parabola opening upwards with vertex at $$(0,-\frac{31}{3})$$.