From the graph provided, we can observe several characteristics of the quadratic function:

1. The graph is a parabola that opens upward, indicating a positive leading coefficient.
2. The vertex is at $(1, -2)$.
3. The parabola crosses the y-axis at $(0, 0)$, meaning the y-intercept is 0.

The standard form of a quadratic equation is:

$y = a(x - h)^2 + k$

Where $(h, k)$ is the vertex of the parabola.

Using the vertex $(1, -2)$, the equation becomes:

$y = a(x - 1)^2 - 2$

Next, we use the point where the parabola intersects the y-axis, $(0, 0)$, to find $a$. Substituting $x = 0$ and $y = 0$ into the equation:

$0 = a(0 - 1)^2 - 2$ $0 = a(1) - 2$ $a = 2$

Thus, the equation of the parabola is:

$y = 2(x - 1)^2 - 2$

Would you like more details on how this equation was derived, or do you have any further questions?

Here are 5 questions to expand the understanding:

1. How can you convert the equation into the standard form $y = ax^2 + bx + c$?
2. What does the value of $a = 2$ tell you about the graph's shape?
3. Can you describe how the vertex form of a quadratic relates to the graph?
4. How would the equation change if the parabola was shifted up or down?
5. What happens to the graph if the $a$ value is negative?

Tip: The vertex form of a quadratic equation is useful for easily identifying the vertex and the direction in which the parabola opens!