Question

nu am mai intalnit functii in fractie pana acuma, si m-a incurcat de tot. Ma poate lamuri cineva cu punctul a si b?

Answer 1

[tex]\\ \boxed{Punctul \:\ A} \\ \\ \\ F_{(x)}= \frac{x^2+6x}{x-2} \\ Aplicam \:\ regula: \left(\frac{f}{g}\right)'=\frac{f'\cdot g-g'\cdot f}{g^2} \\ F`_{(x)}= \frac{\left(x^2+6x\right)`\left(x-2\right)-\left(x-2\right)`\left(x^2+6x\right)}{\left(x-2\right)^2} = \frac{\left(2x+6\right)\left(x-2\right)-1\cdot \left(x^2+6x\right)}{\left(x-2\right)^2} \\ =\ \textgreater \ \frac{x^2-4x-12}{\left(x-2\right)^2} \ \textless \ =\ \textgreater \ \boxed{ \frac{(x-6)(x+2)}{(x-2)^2}} \:\ x \:\ apartine \:\ (2,+infinit) [/tex]



[tex]\\ \boxed{Punctul \:\ B} \\ \\ \\Ecuatia: \boxed{y=mx+n} \:\ unde\:\ \boxed{m= \lim_{+\infty} \frac{F_{(x)}}{x}} \:\ si \:\ \boxed{n= \lim_{+\infty} [F_{(x)}-mx]} \\Acum vine {m= \lim_{+\infty} \frac{\frac{x^2+6x}{x-2}}{x}} =\frac{x^2+6x}{x\left(x-2\right)} = \frac{x+6}{x-2} =\frac{\lim _{x\to \infty \:}\left(1+\frac{6}{x}\right)}{\lim _{x\to \infty \:}\left(1-\frac{2}{x}\right)} =\frac{1}{1} = \boxed{1} [/tex]

[tex]\\ {n= \lim_{+\infty} [\frac{x^2+6x}{x-2}-x]} =\frac{x^2+6x}{x-2}-\frac{x}{1} =\frac{x^2+6x}{x-2}-\frac{x\left(x-2\right)}{x-2}\\ =\frac{x^2+6x}{x-2}-\frac{x\left(x-2\right)}{x-2} =\frac{8x}{x-2} =\ \textgreater \ =\lim _{x\to \infty \:}\left(\frac{8x}{x-2}\right) =8\cdot \frac{\lim _{x\to \infty \:}\left(1\right)}{\lim _{x\to \infty \:}\left(1-\frac{2}{x}\right)} \\ =8\cdot \frac{1}{1} = \boxed{8} \\ y=mx + n =\ \textgreater \ y = 1 \cdot x + 8 =\ \textgreater \ \boxed{y=8x}[/tex]