Question

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question 12 (4 points)
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determine the end behavior for ( f(x) = \frac{x^2 - x}{x^2 - 3x + 2} ).
( \bigcirc ) as ( x \to -infty ), ( f(x) \to 1 )
as ( x \to infty ), ( f(x) \to 1 )
( \bigcirc ) as ( x \to -infty ), ( f(x) \to -1 )
as ( x \to infty ), ( f(x) \to -1 )
( \bigcirc ) as ( x \to -infty ), ( f(x) \to 2 )
as ( x \to infty ), ( f(x) \to 2 )
( \bigcirc ) as ( x \to -infty ), ( f(x) \to 0 )
as ( x \to infty ), ( f(x) \to 0 )
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page 12 of 20

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator.
The numerator \(x^2 - x = x(x - 1)\).
The denominator \(x^2 - 3x + 2=(x - 1)(x - 2)\).
So, \(f(x)=\frac{x(x - 1)}{(x - 1)(x - 2)}\), for \(x
eq1\), we can cancel out \((x - 1)\), and we get \(f(x)=\frac{x}{x - 2}\) (after simplification, note the domain exclusion \(x
eq1\) and \(x
eq2\)).

Step2: Analyze end behavior

To find the end behavior as \(x\to\pm\infty\), we can divide both the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x\) (since the denominator is \(x - 2\), the highest power is \(x^1\)).
For \(f(x)=\frac{x}{x - 2}\), divide numerator and denominator by \(x\):
\(f(x)=\frac{\frac{x}{x}}{\frac{x}{x}-\frac{2}{x}}=\frac{1}{1-\frac{2}{x}}\)
Now, as \(x\to\pm\infty\), \(\frac{2}{x}\to0\) (because as \(x\) becomes very large in magnitude, the fraction \(\frac{2}{x}\) approaches 0).
So, as \(x\to\pm\infty\), \(f(x)=\frac{1}{1 - 0}=1\).

Answer:

As \( x \to -\infty \), \( f(x) \to 1 \); As \( x \to \infty \), \( f(x) \to 1 \) (the first option: "As \( x \to -\infty \), \( f(x) \to 1 \); As \( x \to \infty \), \( f(x) \to 1 \)")