Question

describing the behavior of rational functions
consider the following function:
$f(x) = \frac{2x}{3x^2 - 3}$
which of the following describes the end behavior of $f(x) = \frac{2x}{3x^2 - 3}$? choose two correct answers.

• the graph approaches $\frac{2}{3}$ as $x$ approaches infinity.
• the graph approaches 0 as $x$ approaches infinity.
• the graph approaches 0 as $x$ approaches negative infinity.
• the graph approaches $-1$ as $x$ approaches negative infinity.

Explanation:

Step1: Analyze Degree of Numerator and Denominator

The function is \( f(x)=\frac{2x}{3x^{2}-3} \). The degree of the numerator (\( n \)) is 1, and the degree of the denominator (\( d \)) is 2. Since \( n < d \), as \( x \to \pm\infty \), the function behaves like \( \frac{\text{leading term of numerator}}{\text{leading term of denominator}} \) in terms of end - behavior. But when \( n < d \), the end - behavior is that \( f(x)\to0 \) as \( x\to\pm\infty \).

Step2: Evaluate the Options

• Option 1: "The graph approaches \( \frac{2}{3} \) as \( x \) approaches infinity." This is incorrect. If we were to consider the case when \( n = d \), we would look at the ratio of leading coefficients, but here \( n

eq d \).

• Option 2: "The graph approaches 0 as \( x \) approaches infinity." This is correct because \( n = 1
• Option 3: "The graph approaches 0 as \( x \) approaches negative infinity." This is correct. As \( x\to-\infty \), \( f(x)=\frac{2x}{3x^{2}-3}\approx\frac{2x}{3x^{2}}=\frac{2}{3x}\to0 \) (since \( x\) is negative and its magnitude is large, \( \frac{2}{3x} \) approaches 0).
• Option 4: "The graph approaches - 1 as \( x \) approaches negative infinity." This is incorrect. There is no reason for the function to approach - 1 as \( x\to-\infty \).

Answer:

The two correct answers are:

• The graph approaches 0 as \( x \) approaches infinity.
• The graph approaches 0 as \( x \) approaches negative infinity.