Question

Se considera functia f:R\{-2} -> R, f(x)=$\frac{x^{2}+ax+3 }{x+2}$ , unde a apartine R. Sa se determine a astfel incat graficul functiei sa adminta asimptota oblica dreapta y=x-1 la ramura spre +infinit.

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it y=mx+n\ \ -\ asimptot\breve a\ oblic\breve a\ \ \ \ \ \ (1)\\ \\ y=x-1\ \ -\ asimptot\breve a\ oblic\breve a\ \ \ \ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow m=1,\ \ n=-1\\ \\ -1=n=\lim\limits_{x \to\ \infty}\Big(f(x)-mx\Big)=\lim\limits_{x \to\ \infty}\Big(\dfrac{x^2+ax+3}{x+2}-x\Big)=\\ \\ \\ ==\lim\limits_{x \to\ \infty}\dfrac{x^2+ax+3-x^2-2x}{x+2} ==\lim\limits_{x \to\ \infty}\dfrac{x(a-2)+3}{x+2}=a-2 \Rightarrow\\ \\ \\ a-2=-1 \Rightarrow a=1$