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To find the vertex of the quadratic function \( f(x) = -3x^2 + 6x + 2 \), we can use the vertex formula for a parabola in standard form \( f(x) = ax^2 + bx + c \). The x-coordinate of the vertex is given by the formula:

\[
x = \frac{-b}{2a}
\]

Here, \( a = -3 \), \( b = 6 \), and \( c = 2 \).

Let's calculate the x-coordinate:

\[
x = \frac{-6}{2(-3)} = \frac{-6}{-6} = 1
\]

Now, substitute \( x = 1 \) back into the function to find the y-coordinate:

\[
f(1) = -3(1)^2 + 6(1) + 2 = -3 + 6 + 2 = 5
\]

Thus, the vertex of the function is \( (1, 5) \).

Would you like more details or have any questions?

Here are 5 related questions:
1. How do we identify whether a parabola opens upwards or downwards?
2. What role does the constant \(c\) play in the function?
3. How can we find the axis of symmetry for any quadratic function?
4. How does changing the value of \(a\) affect the shape of the parabola?
5. What is the significance of the vertex in the context of maximum or minimum values?

**Tip:** The vertex is a key feature of a quadratic function, especially when identifying the maximum or minimum value of the function.

Find the vertex of the quadratic function f(x) = -3x^2 + 6x + 2.

Learn how to find the vertex of the quadratic function f(x) = -3x^2 + 6x + 2. This detailed solution explains how to use the vertex formula and substitute the value of x to find the y-coordinate. Suitable for Grades 9-10 students.

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