Question

la punctul a mi a iesit y=1 asimptota orizontala dar la raspunsuri da altceva
la punctul b nu gasesc radacina iar la c la fel​

Answer 1

Functia are valori pozitive oricare ar fi x.

Answer 2

$f(x) = \sqrt{x^2+x+1}-x\\ \\ \text{Pentru }\sqrt{x^2+x+1},\quad x+\dfrac{1}{2}\text{ este asimptota oblica spre }+\infty \\ \\ \text{Deoarece }\Big(x+\dfrac{1}{2}\Big)^2 = x^2+x+\dfrac{1}{4}, \text{ iar }(\sqrt{x^2+x+1})^2 = x^2+x+1 \\ \\ \Rightarrow x^2+x+1\approx x^2+x+\dfrac{1}{4},\quad\text{cand }x\to \infty \\ \\ \text{Pentru }\sqrt{x^2+x+1},\quad -\Big(x+\dfrac{1}{2}\Big)\text{ este asimptota oblica spre }-\infty\\ \\ \Rightarrow \sqrt{x^2+x+1} \approx -\Big(x+\dfrac{1}{2}\Big),\quad x \to -\infty$

$\Rightarrow \lim\limits_{x\to -\infty}\Big(\sqrt{x^2+x+1} -x-mx-n\Big) = \\ \\ =\lim\limits_{x\to -\infty}\Big[-\Big(x+\dfrac{1}{2}\Big)- x-mx-n\Big] = \\ \\ =\lim\limits_{x\to -\infty}\Big( -x-\dfrac{1}{2}-x-mx-n\Big)=\\ \\ =\lim\limits_{x\to -\infty}\left[(-2-m)x-\Big(\dfrac{1}{2}+n\Big) \right] \\ \\\Rightarrow -2-m = 0\Rightarrow m =-2,\quad \dfrac{1}{2}+n =0 \Rightarrow n = -\dfrac{1}{2}$

$\Rightarrow y = mx+n \Rightarrow \boxed{y = -2x-\dfrac{1}{2}}\to \text{asimptota oblica spre } -\infty$