Consider the following rational functions: r(x)=2x−1(x2)−x−2 s(x)=(x3)+27(x2)+4 t(x)=(x3)−9xx+2 u(x)=(x2)+x−6(x2)−25 w(x)=(x3)+(6x2)+9xx+3 Which of these rational functions has a horizontal asymptote?

r(x)=2x−1(x2)−x−2=2x−1(x−2)(x+1) s(x)=(x3)+27(x2)+4=(x+3)(x+9+3x)(x2)+4 t(x)=(x3)−9xx+2=x(x−3)(x+3)x+2 u(x)=(x2)+x−6(x2)−25=(x−2)(x+3)(x−5)(x+5) w(x)=(x3)+(6x2)+9xx+3=x(x+3) r,t and u have horizontal asymptotes

Expert Answer to "Consider the following rational functions: r(x)=2x−1(x2)−x−2 s(x)=(x3)+27(x2)+4 t(x)=(x3)−9xx+2..."

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Alyce Wilkinson

Answered question

2021-09-11

Consider the following rational functions: $r (x) = \frac{2 x - 1}{(x^{2}) - x - 2}$
$s (x) = \frac{(x^{3}) + 27}{(x^{2}) + 4}$
$t (x) = \frac{(x^{3}) - 9 x}{x + 2}$
$u (x) = \frac{(x^{2}) + x - 6}{(x^{2}) - 25}$
$w (x) = \frac{(x^{3}) + (6 x^{2}) + 9 x}{x + 3}$
Which of these rational functions has a horizontal asymptote?

Answer & Explanation

Ezra Herbert

Skilled2021-09-12Added 99 answers

r (x) = \frac{2 x - 1}{(x^{2}) - x - 2} = \frac{2 x - 1}{(x - 2) (x + 1)}

s (x) = \frac{(x^{3}) + 27}{(x^{2}) + 4} = \frac{(x + 3) (x + 9 + 3 x)}{(x^{2}) + 4}

t (x) = \frac{(x^{3}) - 9 x}{x + 2} = \frac{x (x - 3) (x + 3)}{x + 2}

u (x) = \frac{(x^{2}) + x - 6}{(x^{2}) - 25} = \frac{(x - 2) (x + 3)}{(x - 5) (x + 5)}

w (x) = \frac{(x^{3}) + (6 x^{2}) + 9 x}{x + 3} = x (x + 3)

r,t and u have horizontal asymptotes

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