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Let's solve the given problems step by step based on the functions \( s(x) = x - 6 \) and \( t(x) = 3x^2 \).

### 1. \( (s \cdot t)(x) \)
The expression \( (s \cdot t)(x) \) means multiplying the two functions \( s(x) \) and \( t(x) \).
\[
(s \cdot t)(x) = s(x) \cdot t(x) = (x - 6) \cdot (3x^2)
\]
Now expand the product:
\[
(s \cdot t)(x) = (x - 6) \cdot 3x^2 = 3x^3 - 18x^2
\]

### 2. \( (s + t)(x) \)
The expression \( (s + t)(x) \) means adding the two functions \( s(x) \) and \( t(x) \).
\[
(s + t)(x) = s(x) + t(x) = (x - 6) + (3x^2)
\]
So, the result is:
\[
(s + t)(x) = 3x^2 + x - 6
\]

### 3. \( (s - t)(-2) \)
The expression \( (s - t)(x) \) means subtracting the function \( t(x) \) from \( s(x) \), and then evaluating at \( x = -2 \).
\[
(s - t)(x) = s(x) - t(x) = (x - 6) - (3x^2)
\]
Now evaluate at \( x = -2 \):
\[
(s - t)(-2) = (-2 - 6) - (3(-2)^2)
\]
\[
= -8 - 3(4) = -8 - 12 = -20
\]

### Final Answers:
1. \( (s \cdot t)(x) = 3x^3 - 18x^2 \)
2. \( (s + t)(x) = 3x^2 + x - 6 \)
3. \( (s - t)(-2) = -20 \)

Would you like any further details or clarification?

Here are 5 related questions to expand your knowledge:
1. How do you find the derivative of the function \( (s \cdot t)(x) \)?
2. Can you compute the value of \( (s + t)(2) \)?
3. How does the graph of \( s(x) = x - 6 \) compare to that of \( t(x) = 3x^2 \)?
4. What is the inverse of the function \( s(x) = x - 6 \)?
5. How do you solve \( (s - t)(x) = 0 \) for \( x \)?

**Tip:** When multiplying polynomials, always distribute each term carefully to ensure you don't miss any parts!

Suppose that the functions s and t are defined for all real numbers x as follows: s(x) = x - 6 and t(x) = 3x^2. Write the expressions for (s * t)(x) and (s + t)(x) and evaluate (s - t)(-2).

Explore how to compute the product, sum, and difference of the functions s(x) = x - 6 and t(x) = 3x^2. Includes step-by-step solutions for (s * t)(x), (s + t)(x), and the evaluation of (s - t)(-2). Suitable for students in Grades 9-10.

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