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Answer :
We need a polynomial with the following properties for its end behavior:
1. Since as [tex]$x \to +\infty$[/tex] the function decreases (i.e., [tex]$h(x) \to -\infty$[/tex]) and as [tex]$x \to -\infty$[/tex] the function increases (i.e., [tex]$h(x) \to +\infty$[/tex]), the polynomial must be of odd degree.
2. For an odd-degree polynomial to have this behavior, the leading coefficient must be negative. That is, if
[tex]$$ h(x) \sim ax^n \quad \text{with } n \text{ odd}, $$[/tex]
we need [tex]$a < 0$[/tex] so that:
- As [tex]$x \to +\infty$[/tex], [tex]$ax^n \to -\infty$[/tex].
- As [tex]$x \to -\infty$[/tex], [tex]$ax^n \to +\infty$[/tex].
Now, let’s analyze the given options:
- Option A:
[tex]$$ g(x) = 8x^5 - 35x^2 + 20. $$[/tex]
Here, the degree is 5 (which is odd) but the leading coefficient is [tex]$8$[/tex], a positive number. Thus, the end behavior would be:
[tex]$$ \text{As } x \to +\infty,\; g(x) \to +\infty, \quad \text{and as } x \to -\infty,\; g(x) \to -\infty. $$[/tex]
This does not match the required behavior.
- Option B:
[tex]$$ g(x) = 3x^4 - 21x^3 + 12x. $$[/tex]
Here, the degree is 4 (which is even). Even-degree polynomials do not exhibit the alternating behavior required (increasing in one direction and decreasing in the other in this way). Hence, this option is not suitable.
- Option C:
[tex]$$ g(x) = -8x^5 + 15x^2 - 50. $$[/tex]
In this case, the degree is 5 (odd) and the leading coefficient is [tex]$-8$[/tex], which is negative. This means:
[tex]$$ \text{As } x \to +\infty,\; g(x) \to -\infty, \quad \text{and as } x \to -\infty,\; g(x) \to +\infty. $$[/tex]
This exactly matches the end behavior given for [tex]$h(x)$[/tex].
- Option D:
[tex]$$ g(x) = -3x^4 + 18x^3 - 14x. $$[/tex]
The degree here is 4 (even), so it will not show the required end behavior.
Thus, the polynomial that exhibits the same end behavior as [tex]$h(x)$[/tex] is:
[tex]$$ \boxed{-8x^5 + 15x^2 - 50} $$[/tex]
So, the correct answer is Option C.
1. Since as [tex]$x \to +\infty$[/tex] the function decreases (i.e., [tex]$h(x) \to -\infty$[/tex]) and as [tex]$x \to -\infty$[/tex] the function increases (i.e., [tex]$h(x) \to +\infty$[/tex]), the polynomial must be of odd degree.
2. For an odd-degree polynomial to have this behavior, the leading coefficient must be negative. That is, if
[tex]$$ h(x) \sim ax^n \quad \text{with } n \text{ odd}, $$[/tex]
we need [tex]$a < 0$[/tex] so that:
- As [tex]$x \to +\infty$[/tex], [tex]$ax^n \to -\infty$[/tex].
- As [tex]$x \to -\infty$[/tex], [tex]$ax^n \to +\infty$[/tex].
Now, let’s analyze the given options:
- Option A:
[tex]$$ g(x) = 8x^5 - 35x^2 + 20. $$[/tex]
Here, the degree is 5 (which is odd) but the leading coefficient is [tex]$8$[/tex], a positive number. Thus, the end behavior would be:
[tex]$$ \text{As } x \to +\infty,\; g(x) \to +\infty, \quad \text{and as } x \to -\infty,\; g(x) \to -\infty. $$[/tex]
This does not match the required behavior.
- Option B:
[tex]$$ g(x) = 3x^4 - 21x^3 + 12x. $$[/tex]
Here, the degree is 4 (which is even). Even-degree polynomials do not exhibit the alternating behavior required (increasing in one direction and decreasing in the other in this way). Hence, this option is not suitable.
- Option C:
[tex]$$ g(x) = -8x^5 + 15x^2 - 50. $$[/tex]
In this case, the degree is 5 (odd) and the leading coefficient is [tex]$-8$[/tex], which is negative. This means:
[tex]$$ \text{As } x \to +\infty,\; g(x) \to -\infty, \quad \text{and as } x \to -\infty,\; g(x) \to +\infty. $$[/tex]
This exactly matches the end behavior given for [tex]$h(x)$[/tex].
- Option D:
[tex]$$ g(x) = -3x^4 + 18x^3 - 14x. $$[/tex]
The degree here is 4 (even), so it will not show the required end behavior.
Thus, the polynomial that exhibits the same end behavior as [tex]$h(x)$[/tex] is:
[tex]$$ \boxed{-8x^5 + 15x^2 - 50} $$[/tex]
So, the correct answer is Option C.
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