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Let's solve this step by step.

### Question (a)
You are asked to sketch the graph of the equation:
\[ y = (x - 2)(x + 1)(3 - x) \]
and state the intercepts on the coordinate axes.

#### Step 1: Find the x-intercepts
The x-intercepts occur when \( y = 0 \), so we solve the equation:
\[
(x - 2)(x + 1)(3 - x) = 0
\]
This gives us the solutions:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
\[
3 - x = 0 \quad \Rightarrow \quad x = 3
\]
So, the x-intercepts are at \( x = 2 \), \( x = -1 \), and \( x = 3 \).

#### Step 2: Find the y-intercept
The y-intercept occurs when \( x = 0 \). Substituting \( x = 0 \) into the equation:
\[
y = (0 - 2)(0 + 1)(3 - 0) = (-2)(1)(3) = -6
\]
So, the y-intercept is at \( (0, -6) \).

#### Step 3: Sketch the graph
- The curve will pass through the x-intercepts at \( x = -1 \), \( x = 2 \), and \( x = 3 \).
- The y-intercept is at \( (0, -6) \).
- The degree of the equation is cubic (since it is the product of three linear factors), and the leading term is negative (due to \( -x^3 \)), so the graph will start from \( +\infty \) as \( x \to -\infty \), and end at \( -\infty \) as \( x \to +\infty \).

### Question (b)
You are asked to find the values of \( x \) such that:
\[
(x - 2)(x + 1)(3 - x) > 0
\]

We already know the roots are \( x = 2 \), \( x = -1 \), and \( x = 3 \). Now we determine the intervals between and beyond these points where the product is positive.

- The intervals to consider are \( (-\infty, -1) \), \( (-1, 2) \), \( (2, 3) \), and \( (3, \infty) \).
- We test the sign of \( (x - 2)(x + 1)(3 - x) \) in each interval by substituting a value from each interval.

For each interval:
- \( (-\infty, -1) \): Choose \( x = -2 \), then \( (-2 - 2)(-2 + 1)(3 - (-2)) = (-4)(-1)(5) = 20 > 0 \).
- \( (-1, 2) \): Choose \( x = 0 \), then \( (0 - 2)(0 + 1)(3 - 0) = (-2)(1)(3) = -6 < 0 \).
- \( (2, 3) \): Choose \( x = 2.5 \), then \( (2.5 - 2)(2.5 + 1)(3 - 2.5) = (0.5)(3.5)(0.5) = 0.875 > 0 \).
- \( (3, \infty) \): Choose \( x = 4 \), then \( (4 - 2)(4 + 1)(3 - 4) = (2)(5)(-1) = -10 < 0 \).

Thus, the solution to \( (x - 2)(x + 1)(3 - x) > 0 \) is:
\[
x \in (-\infty, -1) \cup (2, 3)
\]

Would you like more details, or do you have any questions?

### Follow-up questions:
1. Can you explain why the cubic graph starts from positive infinity as \( x \to -\infty \)?
2. What happens at the x-intercepts in terms of graph behavior (turning points, crossing the axis)?
3. How can we factor a polynomial if given the roots directly?
4. Why does the graph have a y-intercept of \( -6 \) when \( x = 0 \)?
5. How can the intervals where the product is positive be determined without graphing?

### Tip:
When solving inequalities involving polynomials, testing intervals between the roots will help determine where the expression is positive or negative.

Sketch the graph of y = (x - 2)(x + 1)(3 - x) and find the values of x such that (x - 2)(x + 1)(3 - x) > 0.

Learn how to sketch the graph of y = (x - 2)(x + 1)(3 - x), find the intercepts, and determine where the product of the factors is greater than zero. This is suitable for students in Grades 9-11 and involves concepts in polynomials and inequalities.

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