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.