Question

Consider the following rational functions: r(x)=(2x-1)/(x^2-x-2). s(x)=(x^3+27)/(x^2+4) t(x)=(x^3-9x)/(x+2) u(x)=((x^2)+x-6)/(x^2-25) w(x)=(x^3+6x^2+9x)/(x+3) What are the asymptotes of the function r(x)?

Rational functions
ANSWERED
asked 2021-05-03
Consider the following rational functions: \(\displaystyle{r}{\left({x}\right)}=\frac{{{2}{x}-{1}}}{{{x}^{{2}}-{x}-{2}}}\).
\(\displaystyle{s}{\left({x}\right)}=\frac{{{x}^{{3}}+{27}}}{{{x}^{{2}}+{4}}}\)
\(\displaystyle{t}{\left({x}\right)}=\frac{{{x}^{{3}}-{9}{x}}}{{{x}+{2}}}\)
\(\displaystyle{u}{\left({x}\right)}=\frac{{{\left({x}^{{2}}\right)}+{x}-{6}}}{{{x}^{{2}}-{25}}}\)
\(\displaystyle{w}{\left({x}\right)}=\frac{{{x}^{{3}}+{6}{x}^{{2}}+{9}{x}}}{{{x}+{3}}}\)
What are the asymptotes of the function r(x)?

Answers (1)

2021-05-04
Work as shown below, follow the steps:
\(\displaystyle{r}{\left({x}\right)}=\frac{{{2}{x}-{1}}}{{{x}^{{2}}-{x}-{2}}}\)
- For the vertical asymptotes find the zerosofthe polynomialin the denominator
\(\displaystyle{x}^{{2}}-{x}-{2}={0}\) [write —x as x — 2x]
\(\displaystyle\to{x}^{{2}}+{x}-{2}{x}-{2}={0}\) [group terms 1st-3rd and 2nd-4th]
-> x(x-2)+(x-2)=0 [factor out (x — 2)]
-> (x-2)(x+1)=0 [zero product property]
-> x=2, x=-1 The vertical asymptotes
- For the horizontal asymptote, compare the degrees of the polynomials in the numerator and the denominator:
The degree of the polynomial in the numerator 1 is less than the degree of the polynomial in the denominator 2 so the horizontal asymptote of the graph of this function is the line:
y=0 (the x-axis) The horizontal asymptote
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