Answer to Question #260154 in Calculus for Alunsina

Question

Find the equations of the vertical and horizontal asymptotes of the graphs of the given rational functions.

a. f(x) = 4x²-3x+2/x²-3x+2

b. f(x) = x²-x-20/x²-7x+10

c. f(x) = 4x²-9/2x²-5x+3

d. f(x) = [(4/x-6)+(2/3)]/x

e. f(x) = (4/x²-1)+(2/x+1)

Answer 1

a.

$f(x)=\dfrac{4x^2-3x+2}{x^2-3x+2}$

$x^2-3x+2\not=0=>(x-1)(x-2)\not=0$

Vertical asymptotes: $x=1, x=2.$

$\lim\limits_{x\to-\infin}f(x)=\lim\limits_{x\to-\infin}\dfrac{4x^2-3x+2}{x^2-3x+2}$

$=\lim\limits_{x\to-\infin}\dfrac{4x^2/x^2-3x/x^2+2/x^2}{x^2/x^2-3x/x^2+2/x^2}$

$=\lim\limits_{x\to-\infin}\dfrac{4-3/x+2/x^2}{1-3/x+2/x^2}=\dfrac{4+0+0}{1+0+0}=4$

$\lim\limits_{x\to\infin}f(x)=\lim\limits_{x\to\infin}\dfrac{4x^2-3x+2}{x^2-3x+2}$

$=\lim\limits_{x\to\infin}\dfrac{4x^2/x^2-3x/x^2+2/x^2}{x^2/x^2-3x/x^2+2/x^2}$

$=\lim\limits_{x\to\infin}\dfrac{4-3/x+2/x^2}{1-3/x+2/x^2}=\dfrac{4-0+0}{1-0+0}=4$

Horizontal asymptote: $y=4$

b.

$f(x)=\dfrac{x^2-x-20}{x^2-7x+10}=\dfrac{(x+4)(x-5)}{(x-2)(x-5)}$

Vertical asymptote: $x=2.$

$\lim\limits_{x\to-\infin}f(x)=\lim\limits_{x\to-\infin}\dfrac{x^2-x-20}{x^2-7x+10}$

$=\lim\limits_{x\to-\infin}\dfrac{x^2/x^2-x/x^2-20/x^2}{x^2/x^2-7x/x^2+10/x^2}$

$=\lim\limits_{x\to-\infin}\dfrac{1-1/x-20/x^2}{1-7/x+10/x^2}=\dfrac{1+0-0}{1+0+0}=1$

$\lim\limits_{x\to\infin}f(x)=\lim\limits_{x\to\infin}\dfrac{x^2-x-20}{x^2-7x+10}$

$=\lim\limits_{x\to\infin}\dfrac{x^2/x^2-x/x^2-20/x^2}{x^2/x^2-7x/x^2+10/x^2}$

$=\lim\limits_{x\to\infin}\dfrac{1-1/x-20/x^2}{1-7/x+10/x^2}=\dfrac{1-0-0}{1-0+0}=1$

Horizontal asymptote: $y=1$

c.

$f(x)=\dfrac{4x^2-9}{2x²-5x+3}=\dfrac{(2x+3)(2x-3)}{(x-1)(2x-3)}$

Vertical asymptote: $x=1.$

$\lim\limits_{x\to-\infin}f(x)=\lim\limits_{x\to-\infin}\dfrac{4x^2-9}{2x²-5x+3}$

$=\lim\limits_{x\to-\infin}\dfrac{4x^2/x^2-9/x^2}{2x²/x^2-5x/x^2+3/x^2}$

$=\lim\limits_{x\to-\infin}\dfrac{4-9/x^2}{2-5/x+3/x^2}=\dfrac{4-0}{2+0+0}=2$

$\lim\limits_{x\to\infin}f(x)=\lim\limits_{x\to\infin}\dfrac{4x^2-9}{2x²-5x+3}$

$=\lim\limits_{x\to\infin}\dfrac{4x^2/x^2-9/x^2}{2x²/x^2-5x/x^2+3/x^2}$

$=\lim\limits_{x\to\infin}\dfrac{4-9/x^2}{2-5/x+3/x^2}=\dfrac{4-0}{2-0+0}=2$

d.

$f(x)=\dfrac{\dfrac{4}{x-6}+\dfrac{2}{3}}{x}=\dfrac{12+2x-12}{3x(x-6)}=\dfrac{2x}{3x(x-6)}$

Vertical asymptote: $x=6.$

$\lim\limits_{x\to-\infin}f(x)=\lim\limits_{x\to-\infin}\dfrac{\dfrac{4}{x-6}+\dfrac{2}{3}}{x}$

$=\lim\limits_{x\to-\infin}\dfrac{2}{3(x-6)}=0$

$\lim\limits_{x\to\infin}f(x)=\lim\limits_{x\to\infin}\dfrac{\dfrac{4}{x-6}+\dfrac{2}{3}}{x}$

$=\lim\limits_{x\to\infin}\dfrac{2}{3(x-6)}=0$

Horizontal asymptote: $y=0$

e.

$f(x)=\dfrac{4}{x^2-1}+\dfrac{2}{x+1}=\dfrac{4+2x-2}{(x+1)(x-1)}$

$=\dfrac{2(x+1)}{(x+1)(x-1)}$

Vertical asymptote: $x=1.$

$\lim\limits_{x\to-\infin}f(x)=\lim\limits_{x\to-\infin}(\dfrac{4}{x^2-1}+\dfrac{2}{x+1})$

$=\lim\limits_{x\to-\infin}\dfrac{2(x+1)}{(x+1)(x-1)}=0$

$\lim\limits_{x\to\infin}f(x)=\lim\limits_{x\to\infin}(\dfrac{4}{x^2-1}+\dfrac{2}{x+1})$

$=\lim\limits_{x\to\infin}\dfrac{2(x+1)}{(x+1)(x-1)}=0$

Horizontal asymptote: $y=0$