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Aleks04 [339]
3 years ago
15

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

Mathematics
1 answer:
KatRina [158]3 years ago
4 0

Answer:

<h2>B. x = -1, x = -3, y = 0</h2>

Step-by-step explanation:

g(x)=\dfrac{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}\dfrac{x-2}{x^2+4x+3}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(\frac{1}{x}-\frac{2}{x^2}\right)}{x^2\left(1+\frac{4}{x}+\frac{3}{x^2}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{1}{x}-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{3}{x^2}}=\dfrac{0}{1}=0\\\\\boxed{y=0}

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