Question

consider the following function. ( f(x) = \frac{2x}{3x - 3} ) which of the following describes the end behavior of ( f(x) = \frac{2x}{3x - 3} )? choose two correct answers. the graph approaches 0 as x approaches infinity. the graph approaches -1 as x approaches negative infinity. the graph approaches 0 as x approaches negative infinity. the graph approaches ( \frac{2}{3} ) as x approaches infinity.

Answer:

To determine the end - behavior of the function \(f(x)=\frac{x}{3x - 3}\), we can use the method of dividing both the numerator and the denominator by the highest power of \(x\) in the denominator.

Step 1: Analyze the degrees of the numerator and the denominator

The degree of the numerator (the highest power of \(x\) in the numerator) of the function \(y = \frac{x}{3x-3}\) is \(n = 1\) (since the term is \(x^1\)), and the degree of the denominator (the highest power of \(x\) in the denominator) is also \(m=1\) (the term \(3x\) has power \(1\)).

When the degrees of the numerator and the denominator are equal (\(n = m\)), the end - behavior of the rational function \(y=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\) is determined by the ratio of the leading coefficients. The leading coefficient of the numerator \(a_n = 1\) and the leading coefficient of the denominator \(b_m=3\). So, as \(x\to\pm\infty\), \(y=\frac{x}{3x - 3}=\frac{1}{3-\frac{3}{x}}\to\frac{1}{3}\) (because as \(x\to\pm\infty\), \(\frac{3}{x}\to0\)).

Step 2: Analyze the behavior as \(x\to-\infty\)

We can also rewrite the function \(f(x)=\frac{x}{3x - 3}=\frac{x}{3(x - 1)}\). As \(x\to-\infty\), we consider the sign of the numerator and the denominator. The numerator \(x\to-\infty\) (negative) and the denominator \(3(x - 1)\to3(-\infty- 1)=-\infty\) (negative). A negative number divided by a negative number is positive. But we can also use the limit formula:

\(\lim_{x\to-\infty}\frac{x}{3x - 3}=\lim_{x\to-\infty}\frac{1}{3-\frac{3}{x}}\) (dividing numerator and denominator by \(x\))

Since \(\lim_{x\to-\infty}\frac{3}{x} = 0\), we have \(\lim_{x\to-\infty}\frac{1}{3-\frac{3}{x}}=\frac{1}{3}\). But we can also analyze the function in another way. Let's find the horizontal asymptote. For a rational function \(y = \frac{ax + b}{cx + d}\) (\(a
eq0,c
eq0\)), the horizontal asymptote is \(y=\frac{a}{c}\). Here \(a = 1\), \(c = 3\), so the horizontal asymptote is \(y=\frac{1}{3}\).

Now let's analyze each option:

• Option 1: "The graph approaches 0 as \(x\) approaches infinity" - This is incorrect. As \(x\to\infty\), \(f(x)\to\frac{1}{3}\), not \(0\).
• Option 2: "The graph approaches - 1 as \(x\) approaches negative infinity" - This is incorrect. As \(x\to-\infty\), \(f(x)\to\frac{1}{3}\), not \(- 1\).
• Option 3: "The graph approaches 0 as \(x\) approaches negative infinity" - This is incorrect. As \(x\to-\infty\), \(f(x)\to\frac{1}{3}\), not \(0\).
• Option 4: "The graph approaches \(\frac{1}{3}\) as \(x\) approaches infinity" - This is correct. Because \(\lim_{x\to\infty}\frac{x}{3x - 3}=\lim_{x\to\infty}\frac{1}{3-\frac{3}{x}}=\frac{1}{3}\) (since \(\lim_{x\to\infty}\frac{3}{x}=0\)). Also, when we consider the end - behavior for large \(x\) (positive or negative), the function approaches \(\frac{1}{3}\). Another correct option (assuming there is a mis - typing and one of the options is about approaching \(\frac{1}{3}\) as \(x\) approaches negative infinity) or if we consider the vertical asymptote and other behaviors, but from the given options, the correct ones are:

The correct answers are:

• The graph approaches \(\frac{1}{3}\) as \(x\) approaches infinity.
• (If there is an option like "The graph approaches \(\frac{1}{3}\) as \(x\) approaches negative infinity" which is also correct as we saw from the limit \(\lim_{x\to-\infty}\frac{x}{3x - 3}=\frac{1}{3}\))

If we assume that there is a typo in the options and one of the options is "The graph approaches \(\frac{1}{3}\) as \(x\) approaches negative infinity" and the other is "The graph approaches \(\frac{1}{3}\) as \(x\) approaches infinity", then these two are the correct ones.

If we re - examine the options (assuming the last option is "The graph approaches \(\frac{1}{3}\) as \(x\) approaches infinity") and another correct option (maybe a mis - printed one), but from the limit calculation, the function \(y = \frac{x}{3x-3}\) has a horizontal asymptote \(y=\frac{1}{3}\), so as \(x\to\pm\infty\), \(y\to\frac{1}{3}\).

So the two correct answers are:

• The graph approaches \(\frac{1}{3}\) as \(x\) approaches infinity.
• (If there is an option stating the graph approaches \(\frac{1}{3}\) as \(x\) approaches negative infinity, that is also correct)

If we take the options as given (assuming the last option is "The graph approaches \(\frac{1}{3}\) as \(x\) approaches infinity" and another correct option related to the horizontal asymptote), the correct answers are the ones where the graph approaches \(\frac{1}{3}\) as \(x\) approaches infinity and (if applicable) as \(x\) approaches negative infinity.