Question

Graph the function. y=2x2

Answer 1

The function you're dealing with is \( y = 2x^2 \), which is a quadratic function. Here's a step-by-step guide to understanding and graphing this function:

1. **Identify the Basic Shape**:
- Since the coefficient of \( x^2 \) is positive (2), the parabola opens upwards.

2. **Vertex**:
- The vertex of the parabola \( y = ax^2 + bx + c \) (in standard form) is at the point (0, 0) since there are no linear or constant terms.

3. **Intercepts**:
- **Y-intercept**: Set \( x = 0 \) to find the y-intercept:
\( y = 2(0)^2 = 0 \).
- **X-intercept**: Set \( y = 0 \) to find the x-intercepts:
\( 0 = 2x^2 \) results in \( x = 0 \).
Therefore, the only intercept is at the origin (0, 0).

4. **Other Points**:
- Choose some values for \( x \) to determine \( y \):
- For \( x = 1 \): \( y = 2(1)^2 = 2 \) → point (1, 2)
- For \( x = -1 \): \( y = 2(-1)^2 = 2 \) → point (-1, 2)
- For \( x = 2 \): \( y = 2(2)^2 = 8 \) → point (2, 8)
- For \( x = -2 \): \( y = 2(-2)^2 = 8 \) → point (-2, 8)

5. **Draw the Graph**:
- Plot the points (0, 0), (1, 2), (-1, 2), (2, 8), and (-2, 8) on the graph.
- Sketch a smooth curve through these points to form the parabola.

The graph of the function will look like this:

```
|
| *
| * *
2 | * *
| * *
|* *
|____________________
-3 -2 -1 0 1 2 3
```

The parabola is symmetric about the y-axis, indicating that its left and right sides are mirror images. The vertex at (0,0) is the lowest point of the curve, representing the minimum value of the function.