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The given problem asks for the maximum value of the function:

\[
f(x) = -\frac{2}{3}x^2 + x
\]

### Step-by-Step Solution

1. **Identify the Type of Function:**
   The function \( f(x) = -\frac{2}{3}x^2 + x \) is a quadratic function. Since the coefficient of \( x^2 \) is negative (\(-\frac{2}{3}\)), the parabola opens downwards, indicating that it has a maximum point.

2. **Find the Vertex of the Parabola:**
   For a quadratic function of the form \( ax^2 + bx + c \), the x-coordinate of the vertex (where the maximum or minimum occurs) is given by:

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

   Here, \( a = -\frac{2}{3} \) and \( b = 1 \). Plugging in these values gives:

   \[
   x = -\frac{1}{2 \times \left(-\frac{2}{3}\right)} = \frac{1}{\frac{4}{3}} = \frac{3}{4}
   \]

3. **Calculate the Maximum Value of \( f(x) \):**
   Substitute \( x = \frac{3}{4} \) into the function to find the maximum value:

   \[
   f\left(\frac{3}{4}\right) = -\frac{2}{3} \left( \frac{3}{4} \right)^2 + \frac{3}{4}
   \]

   Compute \( \left( \frac{3}{4} \right)^2 \):

   \[
   \left( \frac{3}{4} \right)^2 = \frac{9}{16}
   \]

   So,

   \[
   f\left(\frac{3}{4}\right) = -\frac{2}{3} \cdot \frac{9}{16} + \frac{3}{4}
   \]

   Simplify the expression:

   \[
   f\left(\frac{3}{4}\right) = -\frac{18}{48} + \frac{36}{48} = \frac{18}{48} = \frac{3}{8}
   \]

4. **Convert to a Common Denominator:**
   To match the choices given, we need to compare:

   \[
   \frac{3}{8} = \frac{9}{24} = \frac{12}{9}
   \]
   
Based on the objective questions

C has the correct. 

Then now we re tell those fine remarks to. 

f(x) = -2/3x^2 + x. Find the maximum value of the function.

Solve the quadratic function f(x) = -2/3x^2 + x and find its maximum value. The solution involves key algebraic concepts such as the vertex formula. Suitable for students in Grades 9-11.

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