match each polynomial with its end behavior as x gets larger and larger in the positive and negative…

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.)\na. ( p(x)=\frac{3}{x - 1} )\nb. ( q(x)=\frac{2x}{x - 1} )\nc. ( r(x)=\frac{2x + 3}{x - 1} )\nd. ( s(x)=\frac{2x^{2}+x + 3}{x - 1} )\ne. ( t(x)=\frac{x^{3}}{x - 1} )\n1. the graph approaches ( y = 2 ).\n2. the graph approaches ( y = 3 ).\n3. the graph approaches ( y = 2x + 3 ).\n4. the graph approaches ( y = x^{2}+x + 1 ).\n5. the graph approaches ( y = 0 ).
Answer
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 (n<m), the horizontal asymptote is (y = 0). If (n=m + 1), there is an oblique asymptote. If (n>m+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<m), (\lim_{x\rightarrow\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\rightarrow\pm\infty}\frac{2x}{x - 1}=\lim_{x\rightarrow\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\rightarrow\pm\infty}\frac{2x + 3}{x - 1}=\lim_{x\rightarrow\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\rightarrow\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\rightarrow\pm\infty}(x^2+x + 1+\frac{1}{x - 1})=x^2+x + 1).
Answer:
- (B)
- (C)
- (D)
- (E)
- (A)