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Hoochie [10]
2 years ago
14

Write the correct answer:

Mathematics
1 answer:
natta225 [31]2 years ago
8 0

Answer:

-1 for question 1, -11 for question 2, 7 for question 3

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for g(x)=x-2/x^2+4x+3
KatRina [158]

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}

4 0
3 years ago
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