Let's break this problem down step by step.

Given Function:

$f(x) = (x + 2)^2(x - 1)(x - 2)$

Part (a): Leading Coefficient and End Behavior

1. Leading Coefficient: To determine the leading coefficient, we need to examine the degree and leading term of the polynomial.

   • The factor $(x + 2)^2$ contributes a degree of 2.
   • The factor $(x - 1)$ contributes a degree of 1.
   • The factor $(x - 2)$ contributes a degree of 1.

   Therefore, the degree of the polynomial is $2 + 1 + 1 = 4$. This is a degree 4 polynomial, and the highest degree term is obtained by multiplying the highest powers of $x$ in each factor.

   • From $(x + 2)^2$, the highest power term is $x^2$.
   • From $(x - 1)$, the highest power term is $x$.
   • From $(x - 2)$, the highest power term is $x$.

   Multiplying these together gives $x^2 \cdot x \cdot x = x^4$.

   The leading term is $x^4$, and the leading coefficient is 1.

2. End Behavior: For a polynomial of degree 4 (even degree) with a positive leading coefficient, the end behavior is: $\text{As } x \to +\infty, \ f(x) \to +\infty \quad (\text{upward})$ $\text{As } x \to -\infty, \ f(x) \to +\infty \quad (\text{upward})$

   Thus, the end behavior is: $\boxed{ \uparrow \quad \text{and} \quad \uparrow }$

Part (b): X-Intercepts, Multiplicity, and Crossing/Touching Behavior

1. X-intercepts: To find the x-intercepts, set $f(x) = 0$.

   $(x + 2)^2(x - 1)(x - 2) = 0$

   This equation is 0 when any of the factors are 0.

   • $x + 2 = 0 \Rightarrow x = -2$
   • $x - 1 = 0 \Rightarrow x = 1$
   • $x - 2 = 0 \Rightarrow x = 2$

   Therefore, the x-intercepts are $x = -2$, $x = 1$, and $x = 2$.

2. Multiplicity and Crossing/Touching Behavior:

   • For $x = -2$, the factor $(x + 2)$ has multiplicity 2 (because it’s squared). When the multiplicity is even, the graph touches the x-axis and turns around at this point.

   • For $x = 1$, the factor $(x - 1)$ has multiplicity 1 (odd). When the multiplicity is odd, the graph crosses the x-axis at this point.

   • For $x = 2$, the factor $(x - 2)$ also has multiplicity 1 (odd). Therefore, the graph crosses the x-axis at this point as well.

Thus, the details are:

• Intercept: $x = -2$, Multiplicity: 2, Cross or Touch: Touch
• Intercept: $x = 1$, Multiplicity: 1, Cross or Touch: Cross
• Intercept: $x = 2$, Multiplicity: 1, Cross or Touch: Cross

Part (c): Y-Intercept

The y-intercept occurs when $x = 0$.

Substitute $x = 0$ into the function: $f(0) = (0 + 2)^2(0 - 1)(0 - 2) = 2^2(-1)(-2) = 4 \cdot (-1) \cdot (-2) = 8$

Thus, the y-intercept is $(0, 8)$.

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Summary:

(a)

• Leading Coefficient: 1
• End Behavior: $\uparrow \quad \text{and} \quad \uparrow$

(b)

• Intercept: $x = -2$, Multiplicity: 2, Touch
• Intercept: $x = 1$, Multiplicity: 1, Cross
• Intercept: $x = 2$, Multiplicity: 1, Cross

(c)

• Y-Intercept: $(0, 8)$

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Would you like to dive into any specific detail, or have more questions? Here are 5 related questions:

1. What happens to the end behavior if the leading coefficient is negative?
2. Can you explain how to find the degree of a polynomial in more detail?
3. How does the multiplicity of an intercept affect the graph's behavior?
4. What is the difference between even and odd degree polynomials in terms of end behavior?
5. How can we determine the sign of the y-intercept without fully solving it?

Tip: The degree and leading coefficient of a polynomial dictate the graph's overall shape and direction.