Let's analyze the given quadratic function:

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

This is a quadratic function in vertex form, $f(x) = a(x-h)^2 + k$, where:

• $a = 5$ (the coefficient in front of the squared term),
• $h = 3$ (the horizontal shift),
• $k = 2$ (the vertical shift).

Key characteristics of the graph:

1. Opens Upward or Downward:

   • Since $a = 5$, which is positive, the parabola opens upward.

2. Vertex:

   • The vertex is at the point $(h, k) = (3, 2)$.

3. Maximum or Minimum:

   • Because the parabola opens upward, it has a minimum value at the vertex, not a maximum.

Thus, the graph opens upward and has a minimum value at $(3, 2)$.

The correct answer:

"Opens upward and has minimum value at (3, 2)."

Would you like further details on this explanation or any other aspects? Here are some questions you might consider:

1. How do we derive the vertex form of a quadratic function?
2. Why does the sign of $a$ determine whether the parabola opens upward or downward?
3. How do we calculate the axis of symmetry from the vertex form?
4. What does changing the value of $k$ (vertical shift) do to the graph of the function?
5. How would the graph change if $a$ were negative?

Tip: In any quadratic equation, the vertex form $f(x) = a(x-h)^2 + k$ is very useful for quickly identifying the direction of the graph and the vertex point.