Question

15 consider the end - behavior description below. as (x\to-infty,f(x)\toinfty) as (x\toinfty,f(x)\to-infty) which function satisfies the given conditions? a (f(x)=x^{4}+2x^{2}+1) c (f(x)=-x^{3}+2x - 6)

Explanation:

Step1: Recall end - behavior rules for polynomials

For a polynomial $f(x)=a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_0$, the end - behavior is determined by the leading term $a_nx^n$. If $n$ is even and $a_n>0$, as $x\to\pm\infty$, $f(x)\to\infty$. If $n$ is even and $a_n < 0$, as $x\to\pm\infty$, $f(x)\to-\infty$. If $n$ is odd and $a_n>0$, as $x\to-\infty$, $f(x)\to-\infty$ and as $x\to\infty$, $f(x)\to\infty$. If $n$ is odd and $a_n < 0$, as $x\to-\infty$, $f(x)\to\infty$ and as $x\to\infty$, $f(x)\to-\infty$.

Step2: Analyze option A

For $f(x)=x^4 + 2x^2+1$, the leading term is $x^4$ (even - degree with $a = 1>0$). So as $x\to\pm\infty$, $f(x)\to\infty$, which does not match the given end - behavior.

Step3: Analyze option B

The graph in option B appears to be a cubic function. But we need to analyze it more precisely.

Step4: Analyze option C

For $f(x)=-x^3 + 2x - 6$, the leading term is $-x^3$. Here, the degree $n = 3$ (odd) and the leading coefficient $a=-1<0$. So as $x\to-\infty$, $f(x)=-(-\infty)^3+2(-\infty)-6=\infty - \infty-6\to\infty$ and as $x\to\infty$, $f(x)=-(\infty)^3+2(\infty)-6=-\infty+\infty - 6\to-\infty$, which matches the given end - behavior.

Step5: Analyze option D

The graph in option D is an even - degree polynomial (since both ends go in the same direction), so it does not match the given end - behavior.

Answer:

C. $f(x)=-x^3 + 2x - 6$