Question

match each polynomial with its end behavior as x gets larger and larger in the positive and negative directions. (note: some of the answer choices are not used and some answer choices are used more than once.)
a. ( p(x)=\frac{3}{x - 1} )
b. ( q(x)=\frac{2x}{x - 1} )
c. ( r(x)=\frac{2x + 3}{x - 1} )
d. ( s(x)=\frac{2x^{2}+x + 3}{x - 1} )
e. ( t(x)=\frac{x^{3}}{x - 1} )

1. the graph approaches ( y = 2 ).
2. the graph approaches ( y = 3 ).
3. the graph approaches ( y = 2x + 3 ).
4. the graph approaches ( y = x^{2}+x + 1 ).
5. the graph approaches ( y = 0 ).

Explanation:

Step1: Analyze the horizontal asymptote formula

For a rational function \(f(x)=\frac{a_nx^n+\cdots}{b_mx^m+\cdots}\), if \(n = m\), the horizontal asymptote is \(y=\frac{a_n}{b_m}\). If \(nm+1\), there is no horizontal or oblique asymptote in the traditional sense.

Step2: Analyze each function

• For \(A: p(x)=\frac{3}{x - 1}\), degree of numerator \(n = 0\), degree of denominator \(m=1\). Since \(n

ightarrow\pm\infty}\frac{3}{x - 1}=0\).

• For \(B: q(x)=\frac{2x}{x - 1}\), degree of numerator \(n = 1\), degree of denominator \(m = 1\). \(\lim_{x

ightarrow\pm\infty}\frac{2x}{x - 1}=\lim_{x
ightarrow\pm\infty}\frac{2}{1-\frac{1}{x}} = 2\).

• For \(C: r(x)=\frac{2x+3}{x - 1}\), degree of numerator \(n = 1\), degree of denominator \(m = 1\). \(\lim_{x

ightarrow\pm\infty}\frac{2x + 3}{x - 1}=\lim_{x
ightarrow\pm\infty}\frac{2+\frac{3}{x}}{1-\frac{1}{x}}=2\).

• For \(D: s(x)=\frac{2x^2+x + 3}{x - 1}\), degree of numerator \(n = 2\), degree of denominator \(m = 1\). Using polynomial long - division: \(\frac{2x^2+x + 3}{x - 1}=2x+3+\frac{6}{x - 1}\). \(\lim_{x

ightarrow\pm\infty}(2x + 3+\frac{6}{x - 1})=2x + 3\).

• For \(E: t(x)=\frac{x^3}{x - 1}\), degree of numerator \(n = 3\), degree of denominator \(m = 1\). Using polynomial long - division: \(\frac{x^3}{x - 1}=x^2+x + 1+\frac{1}{x - 1}\). \(\lim_{x

ightarrow\pm\infty}(x^2+x + 1+\frac{1}{x - 1})=x^2+x + 1\).

Answer:

1. \(B\)
2. \(C\)
3. \(D\)
4. \(E\)
5. \(A\)