Răspuns:
EX1
Asimptote oblice
y=mx+n
m=[tex]\lim_{x \to \infty} \frac{f(x}{x} = \lim_{ \to \infty} \frac{3x^2-x+5}{x(2x+1)}[/tex]
[tex]\lim_{ \to \infty} \frac{3x^2-x+5}{2x^2+x} =\frac{3}{2}[/tex]
n=lim(f(x)-mx)=
[tex]\lim_{x \to \infty}( \frac{3x^2-x+5}{2x+1} -\frac{3x}{2})[/tex]=
[tex]\lim_{x \to \infty} (\frac{6x^2-2x+10}{4x+2} -\frac{6x^2+3x}{4x+2} =[/tex]
[tex]\lim_{x \to \infty}\frac{x+10}{4x+2} =\frac{1}{4}[/tex]
Asimptota Spre +∞
y=[tex]y=\frac{3x}{2} +\frac{1}{4}[/tex]
Analog calculezi asimptota spre -∞
Asimptota verticala
2x+1=0
x=[tex]\frac{-1}{2}[/tex]
Calculezi limita la dreapta si la stanga in [tex]-\frac{1}{2}[/tex]
LS:x<[tex]-\frac{1}{2} \lim_{x \to -\frac{1}{2} } \frac{3x^2-x+5}{2x-1} =[/tex]
[tex]\frac{3*(\frac{-1}{2} )^2-(-\frac{1}{2})+5 }{2*(-\frac{1}{2})+1 } =[/tex][tex]\frac{\frac{3}{4}+\frac{1}{2}+5 }{-1-0+1} =[/tex][tex]\frac{\frac{3+2+20}{4} }{-0} =\frac{25}{-0} =[/tex]-∞
Ld:x>[tex]-\frac{1}{2} \lim_{x \to \inft-\frac{1}{2} } \frac{3x^2-x+5}{2x-1} =[/tex]
[tex]\frac{\frac{25}{4} }{2*(-\frac{1}{2})+0+1 } =[/tex]
[tex]\frac{\frac{25}{4} }{-1+0+1} =[/tex][tex]\frac{25}{+0}=+oo[/tex]
Deci dreapta x=[tex]-\frac{1}{2}[/tex] este asimptota la -∞ pt valori <[tex]-\frac{1}{2}[/tex]si asimptota la +∞ pt valori >[tex]-\frac{1}{2}[/tex]
Explicație pas cu pas:
