We are given a parabola that opens upwards, has a vertex at (0, 0), and passes through the points (-1, 3) and (1, 3). We need to determine which of the given functions matches this graph.

Let's examine each option:

1. Option a: $y = x^2$

   • This is a standard parabola with a vertex at (0, 0).
   • The points (-1, 1) and (1, 1) are on this parabola, but the given points are (-1, 3) and (1, 3). Therefore, this is not the correct equation.

2. Option b: $y = 3x^2$

   • This parabola also has a vertex at (0, 0).
   • For $x = 1$ or $x = -1$, $y = 3(1)^2 = 3$, so the points (-1, 3) and (1, 3) are on this parabola.
   • This matches the given graph.

3. Option c: $y = -\frac{1}{2}x^2$

   • This is a downward-opening parabola due to the negative sign.
   • Since the given parabola opens upwards, this cannot be the correct equation.

4. Option d: $y = \frac{1}{10}x^2$

   • This parabola also opens upwards and has a vertex at (0, 0).
   • However, for $x = 1$ or $x = -1$, $y = \frac{1}{10}(1)^2 = \frac{1}{10}$, which is much smaller than 3. This is not the correct equation.

Conclusion:

The correct function is Option b: $y = 3x^2$.

Would you like further explanation, or do you have any questions?

Here are some related questions:

1. How do you find the vertex of a parabola?
2. What is the formula for a parabola in vertex form?
3. How can you determine if a parabola opens upwards or downwards?
4. How do you find the equation of a parabola given its vertex and a point?
5. What is the difference between a narrow and a wide parabola?

Tip: The coefficient of $x^2$ determines how "narrow" or "wide" a parabola is. The larger the coefficient, the narrower the parabola.