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To determine the end behavior of the polynomial function [tex]\( f(x) = x^2(x-1)(x+3)(x-2) \)[/tex], we will analyze its leading term and degree.
### Step-by-Step Analysis:
1. Expand the Polynomial:
The function [tex]\( f(x) = x^2(x-1)(x+3)(x-2) \)[/tex] consists of products of four factors.
Let's first identify its leading term and degree:
[tex]\[ f(x) = x^2(x-1)(x+3)(x-2) \][/tex]
2. Degree of the Polynomial:
- The degree of [tex]\( x^2 \)[/tex] is 2.
- The degree of [tex]\( (x-1) \)[/tex] is 1.
- The degree of [tex]\( (x+3) \)[/tex] is 1.
- The degree of [tex]\( (x-2) \)[/tex] is 1.
Adding these together gives us the total degree of the polynomial:
[tex]\[ \text{Degree} = 2 + 1 + 1 + 1 = 5 \][/tex]
The polynomial is of degree 5, which means it is an odd-degree polynomial.
3. Leading Term:
When expanded, the leading term comes from the product of the highest degree terms in each of the factors:
[tex]\[ x^2 \cdot x \cdot x \cdot x = x^{5} \][/tex]
So, the leading term is [tex]\( x^5 \)[/tex].
4. End Behavior:
For a polynomial of the form [tex]\( ax^n \)[/tex]:
- If [tex]\( n \)[/tex] is odd, and [tex]\( a \)[/tex] (the leading coefficient) is positive, then as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex], and as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
- If [tex]\( n \)[/tex] is odd and [tex]\( a \)[/tex] is negative, the end behaviors would be reversed.
In our case, the leading coefficient is positive (implicitly 1 in [tex]\( x^5 \)[/tex]) since there are no negative signs involved in the expansion of [tex]\( x^5 \)[/tex].
Therefore, the end behavior follows:
[tex]\[ \begin{cases} \text{As } x \rightarrow \infty, f(x) \rightarrow \infty.\quad (\text{since the leading term } x^5 \text{ dominates})\\ \text{As } x \rightarrow -\infty, f(x) \rightarrow -\infty. \end{cases} \][/tex]
### Conclusion:
The end behaviors of the polynomial function [tex]\( f(x) = x^2(x-1)(x+3)(x-2) \)[/tex] are:
[tex]\[ \boxed{\text{As } x \rightarrow -\infty, f(x) \rightarrow -\infty. \text{ As } x \rightarrow \infty, f(x) \rightarrow \infty.} \][/tex]
### Step-by-Step Analysis:
1. Expand the Polynomial:
The function [tex]\( f(x) = x^2(x-1)(x+3)(x-2) \)[/tex] consists of products of four factors.
Let's first identify its leading term and degree:
[tex]\[ f(x) = x^2(x-1)(x+3)(x-2) \][/tex]
2. Degree of the Polynomial:
- The degree of [tex]\( x^2 \)[/tex] is 2.
- The degree of [tex]\( (x-1) \)[/tex] is 1.
- The degree of [tex]\( (x+3) \)[/tex] is 1.
- The degree of [tex]\( (x-2) \)[/tex] is 1.
Adding these together gives us the total degree of the polynomial:
[tex]\[ \text{Degree} = 2 + 1 + 1 + 1 = 5 \][/tex]
The polynomial is of degree 5, which means it is an odd-degree polynomial.
3. Leading Term:
When expanded, the leading term comes from the product of the highest degree terms in each of the factors:
[tex]\[ x^2 \cdot x \cdot x \cdot x = x^{5} \][/tex]
So, the leading term is [tex]\( x^5 \)[/tex].
4. End Behavior:
For a polynomial of the form [tex]\( ax^n \)[/tex]:
- If [tex]\( n \)[/tex] is odd, and [tex]\( a \)[/tex] (the leading coefficient) is positive, then as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex], and as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
- If [tex]\( n \)[/tex] is odd and [tex]\( a \)[/tex] is negative, the end behaviors would be reversed.
In our case, the leading coefficient is positive (implicitly 1 in [tex]\( x^5 \)[/tex]) since there are no negative signs involved in the expansion of [tex]\( x^5 \)[/tex].
Therefore, the end behavior follows:
[tex]\[ \begin{cases} \text{As } x \rightarrow \infty, f(x) \rightarrow \infty.\quad (\text{since the leading term } x^5 \text{ dominates})\\ \text{As } x \rightarrow -\infty, f(x) \rightarrow -\infty. \end{cases} \][/tex]
### Conclusion:
The end behaviors of the polynomial function [tex]\( f(x) = x^2(x-1)(x+3)(x-2) \)[/tex] are:
[tex]\[ \boxed{\text{As } x \rightarrow -\infty, f(x) \rightarrow -\infty. \text{ As } x \rightarrow \infty, f(x) \rightarrow \infty.} \][/tex]
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