1. **Problem statement:** Sketch the graph of the function $y$ given that $y(1) = 0$ and the graph is a parabola opening upwards with vertex below the x-axis, crossing the x-axis near $x=0$ and $x=2$. 2. **Understanding the parabola:** A parabola opening upwards can be represented by a quadratic function of the form $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex and $a > 0$. 3. **Vertex and roots:** The vertex is below the x-axis, so $k < 0$. The parabola crosses the x-axis near $x=0$ and $x=2$, so the roots are approximately $x=0$ and $x=2$. 4. **Form of the quadratic:** Using roots, the quadratic can be written as $$y = a x (x - 2)$$. 5. **Using the point $y(1) = 0$ to find $a$:** Substitute $x=1$, $y=0$: $$0 = a \cdot 1 \cdot (1 - 2) = a \cdot 1 \cdot (-1) = -a$$ 6. **Solve for $a$:** $$-a = 0 \implies a = 0$$ 7. **Interpretation:** If $a=0$, the function is $y=0$, which contradicts the parabola shape. Since the problem states the parabola shape, the point $y(1)=0$ lies on the parabola between the roots, so the parabola must cross the x-axis at $x=1$ as well. 8. **Adjusting roots:** The parabola crosses the x-axis at $x=0$, $x=1$, and $x=2$ is approximate. Since $y(1)=0$, $x=1$ is a root. 9. **Revised quadratic:** Using roots at $x=0$ and $x=1$, the quadratic is $$y = a x (x - 1)$$. 10. **Vertex location:** The vertex is at $x = \frac{0 + 1}{2} = 0.5$. 11. **Vertex value:** $$y(0.5) = a \cdot 0.5 \cdot (0.5 - 1) = a \cdot 0.5 \cdot (-0.5) = -0.25 a$$ 12. **Since vertex is below x-axis, $k < 0$, so $a > 0$**. 13. **Final sketch:** The parabola opens upwards, passes through $(0,0)$ and $(1,0)$, with vertex at $(0.5, -0.25 a)$ below the x-axis. **Answer:** The graph is a parabola $y = a x (x - 1)$ with $a > 0$ and $y(1) = 0$.