Question

Determine the equations of the vertical and horizontal asymptotes, if any, for g(x)=x-2/x^2+4x+3

A.x=-1,x=3
B.x=-1,x=-3,y=0
C.x=1,x=-3,y=0
D.x=-1,x=-3

Answer 1

B. x = -1, x = -3, y = 0

Explanation:

`g(x)=(x-2)/(x^2+4x+3)\\\\vertical\ asymptote:\\\\x^2+4x+3=0\\x^2+x+3x+3=0\\x(x+1)+3(x+1)=0\\(x+1)(x+3)=0\iff x+1=0\ \vee\ x+3=0\\\\\boxed{x=-1\ \vee\ x=-3}\\\\horizontal\ asymptote:\\\\\lim\limits_(x\to\pm\infty)(x-2)/(x^2+4x+3)=\lim\limits_(x\to\pm\infty)(x^2\left((1)/(x)-(2)/(x^2)\right))/(x^2\left(1+(4)/(x)+(3)/(x^2)\right))=\lim\limits_(x\to\pm\infty)((1)/(x)-(2)/(x^2))/(1+(4)/(x)+(3)/(x^2))=(0)/(1)=0\\\\\boxed{y=0}`