Question

Mă puteți ajuta să rezolv aceste Asimtote?

Answer 1

Răspuns:

EX1

Asimptote oblice

y=mx+n

m=$\lim_{x \to \infty} \frac{f(x}{x} = \lim_{ \to \infty} \frac{3x^2-x+5}{x(2x+1)}$

$\lim_{ \to \infty} \frac{3x^2-x+5}{2x^2+x} =\frac{3}{2}$

n=lim(f(x)-mx)=

$\lim_{x \to \infty}( \frac{3x^2-x+5}{2x+1} -\frac{3x}{2})$=

$\lim_{x \to \infty} (\frac{6x^2-2x+10}{4x+2} -\frac{6x^2+3x}{4x+2} =$

$\lim_{x \to \infty}\frac{x+10}{4x+2} =\frac{1}{4}$

Asimptota Spre +∞

y=$y=\frac{3x}{2} +\frac{1}{4}$

Analog calculezi asimptota spre -∞

Asimptota verticala

2x+1=0

x=$\frac{-1}{2}$

Calculezi limita  la  dreapta si la   stanga in $-\frac{1}{2}$

LS:x<$-\frac{1}{2} \lim_{x \to -\frac{1}{2} } \frac{3x^2-x+5}{2x-1} =$

$\frac{3*(\frac{-1}{2} )^2-(-\frac{1}{2})+5 }{2*(-\frac{1}{2})+1 } =$$\frac{\frac{3}{4}+\frac{1}{2}+5 }{-1-0+1} =$$\frac{\frac{3+2+20}{4} }{-0} =\frac{25}{-0} =$-∞

Ld:x>$-\frac{1}{2} \lim_{x \to \inft-\frac{1}{2} } \frac{3x^2-x+5}{2x-1} =$

$\frac{\frac{25}{4} }{2*(-\frac{1}{2})+0+1 } =$

$\frac{\frac{25}{4} }{-1+0+1} =$$\frac{25}{+0}=+oo$

Deci dreapta x=$-\frac{1}{2}$ este asimptota la -∞ pt   valori <$-\frac{1}{2}$si asimptota  la +∞ pt valori >$-\frac{1}{2}$

Explicație pas cu pas: