Question

consider the end behavior of the function $g(x) = -\frac{1}{3}|x| + 9$ and complete the statements. as $x$ approaches negative infinity, $g(x)$ approaches $\boldsymbol{\text{drop - down}}$ infinity. as $x$ approaches positive infinity, $g(x)$ approaches $\boldsymbol{\text{drop - down}}$ infinity.

Explanation:

Step1: Analyze |x| as x→-∞

As $x \to -\infty$, $|x| = -x$, so $|x| \to +\infty$.

Step2: Compute g(x) as x→-∞

$g(x) = -\frac{1}{3}|x| + 9$, so $-\frac{1}{3}|x| \to -\infty$, thus $g(x) \to -\infty$.

Step3: Analyze |x| as x→+∞

As $x \to +\infty$, $|x| = x$, so $|x| \to +\infty$.

Step4: Compute g(x) as x→+∞

$g(x) = -\frac{1}{3}|x| + 9$, so $-\frac{1}{3}|x| \to -\infty$, thus $g(x) \to -\infty$.

Answer:

As x approaches negative Infinity, $g(x)$ approaches negative Infinity.
As x approaches positive Infinity, $g(x)$ approaches negative Infinity.