Odpowiedź:
Asymptota pionowa:
[tex]\huge\boxed{x=1\ \wedge\ x=4}[/tex]
Asymptota ukośna:
[tex]\huge\boxed{y=3x+5}[/tex]
Szczegółowe wyjaśnienie:
Asymptota pozioma:
[tex]\lim\limits_{x\to\pm\infty}g(x)=\lim\limits_{x\to\pm\infty}\dfrac{3x^3-10x^2-3x+12}{x^2-5x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^3\left(30-\frac{10}{x}-\frac{3}{x^2}+\frac{12}{x^3}\right)}{x^2\left(1-\frac{5}{x}+\frac{4}{x^2}\right)}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x\left(30-\frac{10}{x}-\frac{3}{x^2}+\frac{12}{x}\right)}{1-\frac{5}{x}+\frac{4}{x^2}}=\pm\infty[/tex]
nie istnieje
Asymptota pionowa:
[tex]D:x^2-5x+4\neq0\\\\x^2-x-4x+4\neq0\\\\x(x-1)-4(x-1)\neq0\\\\(x-1)(x-4)\neq0\iff x-1\neq0\ \wedge\ x-4\neq0\\\\x=1\ \wedge\ x\neq4[/tex]
Dwie asymptoty pionowe [tex]x=1[/tex] oraz [tex]x=4[/tex]
Asymptoty ukośne:
[tex]y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{g(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}\left(g(x)-ax\right)[/tex]
[tex]\lim\limits_{x\to\pm\infty}\dfrac{\frac{3x^3-10x^2-3x+12}{x^2-5x+4}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{3x^3-10x^2-3x+12}{x^3-5x^2+4x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^3\!\!\!\!\!\!\diagup\left(3-\frac{10}{x}-\frac{3}{x^2}+\frac{12}{x}\right)}{x^3\!\!\!\!\!\!\diagup\left(1-\frac{5}{x}+\frac{4}{x^2}\right)}=\dfrac{3}{1}=3[/tex]
stąd mamy [tex]a=3[/tex]
Obliczamy [tex]b[/tex]:
[tex]\lim\limits_{x\to\pm\infty}\left(\dfrac{3x^3-10x^2-3z+12}{x^2-5x+4}-3x\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{3x^3-10x^2-3x+12}{x^2-5x+4}-\dfrac{3x(x^2-5x+4)}{x^2-5x+4}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{3x^3-10x^2-3x+12-3x^3+15x^2-12x}{x^2-5x+4}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{5x^2-15x+12}{x^2-5x+4}\right)[/tex]
[tex]=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2\!\!\!\!\!\!\diagup\left(5-\frac{15}{x}+\frac{12}{x^2}\right)}{x^2\!\!\!\!\!\!\diagup\left(1-\frac{5}{x}+\frac{4}{x^2}\right)}\right)\\\\=\lim\limits_{x\to\pm\infty}\dfrac{5-\frac{15}{x}+\frac{12}{x^2}}{1-\frac{5}{x}+\frac{4}{x^2}}=\dfrac{5}{1}=5[/tex]
stąd mamy [tex]b=5[/tex]
Równanie asymptoty: [tex]y=3x+5[/tex]
