Vertex is (0,2) a > 0 didn't I give you this

It looks like you're working with a quadratic function, and you've provided some details. Here's what I can infer from the information:

- The vertex of the parabola is at \((0, 2)\).
- Since \(a > 0\), the parabola opens upwards.

Given this, the general form of a quadratic function is:

\[
y = a(x - h)^2 + k
\]

where \((h, k)\) is the vertex of the parabola. Since the vertex is \((0, 2)\), we can substitute \(h = 0\) and \(k = 2\):

\[
y = a(x - 0)^2 + 2
\]

or simplified:

\[
y = ax^2 + 2
\]

Since \(a > 0\), the parabola will open upwards, and the vertex at \((0, 2)\) will be the minimum point.

Would you like to explore more details, such as finding \(a\) if additional points are given?

Here are some related questions:
1. How do you find \(a\) if another point on the parabola is provided?
2. What is the axis of symmetry of this parabola?
3. How would the graph change if \(a < 0\)?
4. How can you determine the direction of a parabola from its equation?
5. How does the vertex form differ from the standard form of a quadratic equation?

**Tip:** To determine the shape of the parabola, knowing just \(a\) (positive or negative) can tell you if it opens upwards or downwards!

Explore how to represent a quadratic function with a vertex at (0, 2) and a positive leading coefficient (a > 0). Learn the vertex form of quadratic equations, key algebraic concepts, and properties of parabolas. Ideal for Grades 9-10 students.

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