1. The problem is to analyze the equation $X = y^2$. 2. This equation represents a parabola opening to the right in the $xy$-plane. 3. The general form of a parabola that opens horizontally is $X = ay^2 + by + c$. Here, $a=1$, $b=0$, and $c=0$. 4. To understand the shape, note that for each value of $y$, $X$ is the square of $y$, so $X$ is always non-negative. 5. The vertex of the parabola is at the origin $(0,0)$. 6. The parabola is symmetric about the $X$-axis because $X$ depends on $y^2$. 7. The intercept with the $X$-axis is at $X=0$ when $y=0$. 8. There are no $Y$-intercepts because $X$ cannot be negative. 9. The function can be rewritten as $y = \pm \sqrt{X}$ to express $y$ in terms of $X$. 10. This is a standard parabola with vertex at the origin and opening to the right. Final answer: The graph of $X = y^2$ is a parabola opening to the right with vertex at $(0,0)$.