Question

based on the table, which best predicts the end behavior of the graph of $f(x)$? the table has two rows: $x$ and $f(x)$. the values of $x$ are -5, -4, -3, -2, -1, 0, 1, 2, 3. the corresponding values of $f(x)$ are -6, -2, 0, 4, 4, 0, -2, -6, -10. the options are: 1. as $x \to \infty$, $f(x) \to \infty$, and as $x \to -\infty$, $f(x) \to -\infty$. 2. as $x \to \infty$, $f(x) \to -\infty$, and as $x \to -\infty$, $f(x) \to \infty$. 3. as $x \to \infty$, $f(x) \to \infty$, and as $x \to -\infty$, $f(x) \to \infty$. 4. as $x \to \infty$, $f(x) \to -\infty$, and as $x \to -\infty$, $f(x) \to \infty$.

Explanation:

Step1: Identify degree and leading coefficient

First, we test if the function is a polynomial. The finite differences:
1st differences: $-6-(-10)=4$, $-2-(-6)=4$, $0-(-2)=2$, $4-0=4$, $4-4=0$, $0-4=-4$, $-2-0=-2$, $-6-(-2)=-4$
2nd differences: $4-4=0$, $2-4=-2$, $4-2=2$, $0-4=-4$, $-4-0=-4$, $-2-(-4)=2$, $-4-(-2)=-2$
3rd differences: $-2-0=-2$, $2-(-2)=4$, $-4-2=-6$, $-4-(-4)=0$, $2-(-4)=6$, $-2-2=-4$
4th differences: $4-(-2)=6$, $-6-4=-10$, $0-(-6)=6$, $6-0=6$, $-4-6=-10$
5th differences: $-10-6=-16$, $6-(-10)=16$, $6-6=0$, $-10-6=-16$
6th differences: $16-(-16)=32$, $0-16=-16$, $-16-0=-16$
Since differences are not constant, observe end behavior trend:
As $x$ increases to $3$, $f(x)=-10$ (decreasing to $-\infty$). As $x$ decreases to $-5$, $f(x)=-6$ (decreasing to $-\infty$). This matches an even-degree polynomial with negative leading coefficient: as $x\to\pm\infty$, $f(x)\to-\infty$.

Step2: Match to options

Compare the trend to the given choices. The option stating "As $x \to \infty, f(x) \to -\infty$, and as $x \to -\infty$, $f(x) \to -\infty$" aligns with the observed trend.

Answer:

As $x \to \infty, f(x) \to -\infty$, and as $x \to -\infty$, $f(x) \to -\infty$