https://file.aiquickdraw.com/mathbot/file/1729038301866br4vvjyp.jpg

Let's break down the problem in the image step by step.

We are given the relation:
\[
-2x^2 + 2x + 3y = -3y
\]

Our goal is to evaluate the function found in the previous step at \(x = 2\).

### Step 1: Solve for \(y\)

We need to first rearrange the given relation to express \(y\) as a function of \(x\). Start by simplifying the equation:
\[
-2x^2 + 2x + 3y = -3y
\]

Add \(3y\) to both sides:
\[
-2x^2 + 2x + 3y + 3y = 0
\]
\[
-2x^2 + 2x + 6y = 0
\]

Now, solve for \(y\):
\[
6y = 2x^2 - 2x
\]
\[
y = \frac{2x^2 - 2x}{6}
\]
\[
y = \frac{x^2 - x}{3}
\]

Thus, the function we have is:
\[
y = \frac{x^2 - x}{3}
\]

### Step 2: Evaluate \(y\) at \(x = 2\)

Substitute \(x = 2\) into the function:
\[
y(2) = \frac{2^2 - 2}{3} = \frac{4 - 2}{3} = \frac{2}{3}
\]

Thus, \( f(2) = \frac{2}{3} \).

### Final Answer:
\[
f(2) = \frac{2}{3}
\]

Would you like further details or have any questions on this?

Here are five related questions to deepen your understanding:
1. How would the function change if we added a constant to both sides of the equation?
2. What happens if you evaluate the function at a negative \(x\) value?
3. Can you derive an alternative form of the function if the original equation had a different coefficient for \(y\)?
4. How does the degree of the polynomial affect the shape of the function graph?
5. What would the result be for other values of \(x\), like \(x = 0\) or \(x = -1\)?

**Tip**: Always simplify the equation step-by-step when solving for a variable to avoid mistakes with coefficients.

Evaluate the function found in the previous step at x = 2 using the relation -2x^2 + 2x + 3y = -3y.

Learn how to evaluate a function derived from the quadratic equation -2x^2 + 2x + 3y = -3y by substituting x = 2. This problem explores algebraic manipulation, rearranging equations, and solving quadratic expressions. Suitable for students in Grades 9-12.

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