1. **State the problem:** We have a family of parabolas defined by a general equation involving $n$, $x$, and $y$. For $n=1,2,3$, the parabolas are graphed, all opening upward with vertex near $(0,1)$. We need to find which equation among the options fits all $n \geq 1$. 2. **Analyze the vertex:** The vertex is near $(0,1)$ for all parabolas, so the equation should have a constant term $+1$. 3. **Analyze the shape and width:** The parabola for $n=1$ is the narrowest and closest to the y-axis, $n=2$ is wider, and $n=3$ is the widest. This means as $n$ increases, the parabola becomes wider (less steep). 4. **Check each option:** - A. $y = nx^2 + 1$: As $n$ increases, the parabola becomes steeper (narrower), which contradicts the graph. - B. $y = \frac{1}{n}x^2 + 1$: As $n$ increases, $\frac{1}{n}$ decreases, so the parabola becomes wider, matching the graph. - C. $y = x^2 + n$: Vertex moves up as $n$ increases, contradicting vertex at $(0,1)$. - D. $y = -nx^2 + 1$: Parabolas open downward, contradicting upward opening. - E. $y = -\frac{1}{n}x^2 + 1$: Parabolas open downward, contradicting upward opening. 5. **Conclusion:** The correct general equation is $$y = \frac{1}{n}x^2 + 1$$ which is option B. **Final answer:** B