Question: Graph the parabola.
y = \frac{1}{4} x^2
Plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. Then click on the graph-a-function button.
Graph shape: a parabola opening upward with vertex at (0, 0), position_hint = center.
1. **State the problem:** We need to graph the parabola given by the equation $$y = \frac{1}{4} x^2$$ and plot five points: the vertex, two points to the left, and two points to the right of the vertex.
2. **Identify the vertex:** The vertex of the parabola $$y = a x^2$$ is at the point $$(0,0)$$ since there is no horizontal or vertical shift.
3. **Choose points to the left and right of the vertex:** We select $x = -2, -1, 1, 2$ to find corresponding $y$ values.
4. **Calculate the $y$ values:**
- For $x = -2$: $$y = \frac{1}{4} (-2)^2 = \frac{1}{4} \times 4 = 1$$
- For $x = -1$: $$y = \frac{1}{4} (-1)^2 = \frac{1}{4} \times 1 = \frac{1}{4}$$
- For $x = 1$: $$y = \frac{1}{4} (1)^2 = \frac{1}{4} \times 1 = \frac{1}{4}$$
- For $x = 2$: $$y = \frac{1}{4} (2)^2 = \frac{1}{4} \times 4 = 1$$
5. **List the five points:**
- Vertex: $$(0,0)$$
- Left points: $$(-2,1), (-1, \frac{1}{4})$$
- Right points: $$(1, \frac{1}{4}), (2,1)$$
6. **Graph shape:** The parabola opens upward because the coefficient $$\frac{1}{4}$$ is positive.
7. **Summary:** The parabola $$y = \frac{1}{4} x^2$$ has vertex at $$(0,0)$$ and passes through points $$(-2,1), (-1, \frac{1}{4}), (1, \frac{1}{4}), (2,1)$$.
These points can be plotted to sketch the parabola accurately.