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To solve the given problem, we need to find \((g \circ f)(-2 + x)\), which means we first need to compute \(f(-2 + x)\) and then use that result as the input for \(g(x)\).

### Given functions:
- \(g(x) = 2x - 2\)
- \(f(x) = x^2 + 3x\)

### Step 1: Find \(f(-2 + x)\)
We substitute \((-2 + x)\) into the function \(f(x)\):

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

Expanding this expression:

\[
(-2 + x)^2 = 4 - 4x + x^2
\]
\[
3(-2 + x) = -6 + 3x
\]

Now, combine both parts:

\[
f(-2 + x) = (4 - 4x + x^2) + (-6 + 3x)
\]
\[
f(-2 + x) = x^2 - x - 2
\]

### Step 2: Use this result in \(g(x)\)

Now that we have \(f(-2 + x) = x^2 - x - 2\), we substitute this into \(g(x)\):

\[
g(f(-2 + x)) = g(x^2 - x - 2)
\]
Substitute into \(g(x)\):

\[
g(x^2 - x - 2) = 2(x^2 - x - 2) - 2
\]
Expanding this:

\[
g(x^2 - x - 2) = 2x^2 - 2x - 4 - 2
\]
\[
g(x^2 - x - 2) = 2x^2 - 2x - 6
\]

Thus, the final answer is:

\[
(g \circ f)(-2 + x) = 2x^2 - 2x - 6
\]

Would you like any more details or explanations? Here are five related questions to deepen your understanding:

1. What is the composition of two functions and how does it work in general?
2. How do you expand a binomial like \((-2 + x)^2\)?
3. What happens when you compose two functions in the reverse order, such as \((f \circ g)(x)\)?
4. Can the process of function composition change the domain of the resulting function?
5. What are the potential real-life applications of function composition?

**Tip:** Always be mindful of the order when composing functions; mixing up the order can change the result!

Find (g ∘ f)(-2 + x) where g(x) = 2x - 2 and f(x) = x^2 + 3x.

Learn how to solve the composition of functions (g ∘ f)(-2 + x) with given functions g(x) = 2x - 2 and f(x) = x^2 + 3x. This step-by-step guide covers function composition, polynomial expansion, and algebraic manipulation. Suitable for students in Grades 9-11.

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