1. The problem is to graph a quadratic function that produces a U-shaped curve with a vertex (lowest point) around $y=6$ and x-values roughly between $-1.5$ and $1.5$. 2. The general form of a quadratic function is $$y = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. 3. For a U-shaped curve (a parabola opening upwards), the coefficient $a$ must be positive. 4. The vertex form of a quadratic function is $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex of the parabola. 5. Since the lowest point is around $y=6$, we set $k=6$. 6. The vertex is near the center, so $h=0$ approximately. 7. Thus, the function can be written as $$y = a x^2 + 6$$. 8. To find $a$, use the approximate x-range where the curve reaches $y=20$ at $x=\pm 1.5$: $$20 = a(1.5)^2 + 6$$ $$20 - 6 = a \times 2.25$$ $$14 = 2.25a$$ $$a = \frac{14}{2.25} = \frac{56}{9} \approx 6.22$$ 9. So the quadratic function is approximately $$y = 6.22 x^2 + 6$$. 10. Plotting this function will give a U-shaped curve with vertex at $(0,6)$ and y-values reaching about 20 at $x=\pm 1.5$. This matches the description of the graph you want.