Question

f(x) = \frac{(x - 2)(x + 1)}{x + 1}
which statements describe the end behavior of the graph of the function shown? check all that apply.
\square as x \to \infty, y \to 1.
\square as x \to \infty, y \to \infty.
\square as x \to \infty, y \to -\infty.
\square as x \to -\infty, y \to 1.
\square as x \to -\infty, y \to \infty.
\square as x \to -\infty, y \to -\infty.

Explanation:

Step1: Simplify the function

First, we simplify the given function \( f(x)=\frac{(x - 2)(x + 1)}{x + 1} \). For \( x
eq - 1 \), we can cancel out the \( x + 1 \) terms, so \( f(x)=x - 2 \) (with a hole at \( x=-1 \)).

Step2: Analyze end - behavior of the simplified function

The function \( y=x - 2 \) is a linear function with a slope of \( m = 1 \) and a y - intercept of \( b=-2 \). The end - behavior of a linear function \( y=mx + b \) (where \( m
eq0 \)) is determined by the slope.

• As \( x

ightarrow\infty \):
We use the formula for the limit of a linear function. For \( y=x - 2 \), \( \lim_{x
ightarrow\infty}(x - 2)=\lim_{x
ightarrow\infty}x-\lim_{x
ightarrow\infty}2 \). Since \( \lim_{x
ightarrow\infty}x=\infty \) and \( \lim_{x
ightarrow\infty}2 = 2 \), but wait, there is a mistake in the previous simplification interpretation. Wait, no, the original function after canceling \( x + 1 \) (for \( x
eq - 1 \)) is \( f(x)=x - 2 \), but the end - behavior of a linear function \( y = x-2 \) has \( \lim_{x
ightarrow\infty}(x - 2)=\infty \)? Wait, no, this is wrong. Wait, no, the user's function is \( f(x)=\frac{(x - 2)(x + 1)}{x + 1} \). Let's expand the numerator: \( (x - 2)(x + 1)=x^{2}+x-2x - 2=x^{2}-x - 2 \). Then \( f(x)=\frac{x^{2}-x - 2}{x + 1} \). We can perform polynomial long division or rewrite the numerator:
\( x^{2}-x - 2=(x + 1)(x-2) \), so for \( x
eq - 1 \), \( f(x)=x - 2 \). But when we consider the end - behavior, we can also think about the original rational function. The degree of the numerator \( n = 2 \) and the degree of the denominator \( d = 1 \). Since \( n>d \), the end - behavior is determined by the leading terms. The leading term of the numerator is \( x^{2} \) and the leading term of the denominator is \( x \). So \( \frac{x^{2}}{x}=x \). But when we cancel the \( x + 1 \) terms (for \( x
eq - 1 \)), we get a linear function. Wait, there is a contradiction here. Wait, no, \( (x - 2)(x + 1)=x^{2}-x - 2 \), so the numerator is degree 2, denominator is degree 1. But when we cancel \( x + 1 \), we are left with a linear function. The key is that the original function has a removable discontinuity at \( x=-1 \), and for the rest of the domain (\( x
eq - 1 \)), it is equivalent to \( y=x - 2 \).
The linear function \( y=x - 2 \) has a slope of 1. So:

• As \( x

ightarrow\infty \), \( y=x - 2
ightarrow\infty \)

• As \( x

ightarrow-\infty \), \( y=x - 2
ightarrow-\infty \)

Wait, but this contradicts the initial thought. Wait, no, the user's function: let's re - evaluate. The function \( f(x)=\frac{(x - 2)(x + 1)}{x + 1} \) is equal to \( x - 2 \) for \( x
eq - 1 \). So it's a linear function with a hole at \( x=-1 \). The end - behavior of a linear function \( y = mx + b \) (here \( m = 1 \), \( b=-2 \)):

• When \( x

ightarrow\infty \), since the slope \( m = 1>0 \), \( y=x - 2
ightarrow\infty \)

• When \( x

ightarrow-\infty \), since the slope \( m = 1>0 \), \( y=x - 2
ightarrow-\infty \)

But wait, the options given are:

1. As \( x

ightarrow\infty,y
ightarrow1 \)

1. As \( x

ightarrow\infty,y
ightarrow\infty \)

1. As \( x

ightarrow\infty,y
ightarrow-\infty \)

1. As \( x

ightarrow-\infty,y
ightarrow1 \)

1. As \( x

ightarrow-\infty,y
ightarrow\infty \)

1. As \( x

ightarrow-\infty,y
ightarrow-\infty \)

From our analysis of \( f(x)=x - 2 \) (for \( x
eq - 1 \)):

• As \( x

ightarrow\infty \), \( y=x - 2
ightarrow\infty \) (so the second option is correct)

• As \( x

ightarrow-\infty \), \( y=x - 2
ightarrow-\infty \) (so the sixth option is correct)

Answer:

As \( x
ightarrow\infty,y
ightarrow\infty \) (the option "As \( x
ightarrow\infty,y
ightarrow\infty \)") and As \( x
ightarrow-\infty,y
ightarrow-\infty \) (the option "As \( x
ightarrow-\infty,y
ightarrow-\infty \)") are the correct ones. If we need to box the options (assuming the options are labeled as follows: let the first option be A, second B, third C, fourth D, fifth E, sixth F):
B. As \( x
ightarrow\infty,y
ightarrow\infty \)
F. As \( x
ightarrow-\infty,y
ightarrow-\infty \)