The graph appears to show a quadratic function with a vertex at (2, 5) and passing through the point (3, 2). The general form of a quadratic equation is:

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

where $(h, k)$ is the vertex of the parabola. Based on the graph, the vertex is $(2, 5)$, so we substitute $h = 2$ and $k = 5$:

$f(x) = a(x - 2)^2 + 5$

Now, we use the point $(3, 2)$ to solve for $a$. Substituting $x = 3$ and $f(x) = 2$ into the equation:

$2 = a(3 - 2)^2 + 5$

Simplifying:

$2 = a(1)^2 + 5$ $2 = a + 5$ $a = 2 - 5 = -3$

Thus, the equation of the quadratic function is:

$f(x) = -3(x - 2)^2 + 5$

Let me know if you need further details or clarifications!

Here are some related questions for practice:

1. How would the equation change if the vertex moved to (1, -3)?
2. What is the role of the coefficient $a$ in determining the parabola's direction?
3. How can you derive the equation of a parabola given its roots and vertex?
4. How does the vertex form differ from the standard form $ax^2 + bx + c$?
5. How can the axis of symmetry of a parabola be determined from its equation?

Tip: The sign of $a$ determines whether the parabola opens upwards or downwards.