Question

Алгебра.................... ​

Answer 1

Ответ:

2)

$y=\dfrac{ax+4}{2x+b}\ \ ,\ \ \ x=2\ ,\ y=1\ \ -\ \ asimptotu$

$y=1\ -\ asimptota\ \ \Rightarrow \ \ \lim\limits _{x \to \infty}\dfrac{ax+4}{2x+b}= \lim\limits _{x \to \infty}\dfrac{a+\frac{4}{x}}{2+\frac{b}{x}}=\dfrac{a}{2}=1\ \ \Rightarrow \\\\a=2\\\\ x=2\ -\ asimptota\ \ \Rightarrow \ \ \lim\limits _{x \to 2}\dfrac{2x+4}{2x+b}=\infty \ \ \Rightarrow \ \ \dfrac{2\cdot 2+4}{2\cdot 2+b}=\infty \ \ \Rightarrow \\\\4+b=0\ \ \ \Rightarrow \ \ \ b=-4$

$y=\dfrac{2x+4}{2x-4}=\dfrac{x+2}{x-2}=\dfrac{(x-2)+2+2}{x-2}=1+\dfrac{4}{x-2}\\\\\\n+\dfrac{k}{x+m}=1+\dfrac{4}{x-2}\ \ \ \Longrightarrow \ \ \ \ \underline {\ n=1\ ,\ k=4\ ,\ m=-2\ }\\\\\\\boxed {\ y=1+\dfrac{4}{x-2}\ }$

$3)\ \ y=\dfrac{3}{x+5}\\\\x+5=\dfrac{3}{y}\ \ ,\ \ \ x=\dfrac{3}{y}-5\ \ \ \Rightarrow \ \ \ x=\dfrac{3-5y}{y}\ \ ,\ \ \ \boxed {\ y=\dfrac{3-5x}{x}\ }\ -\ obratnaya$

$4)\ \ f(x)=\dfrac{3}{x+2}\ \ ,\ \ g(x)=3x+4\\\\\\z(x)=g(f(x))=3g(x)+4=3\cdot \dfrac{3}{x+2}+4=\dfrac{9+4x+8}{x+2}=\dfrac{4x+17}{x+2}\\\\\\\boxed{\ z(x)=\dfrac{4x+17}{x+2}\ }$