Find the all asymtotes of the curve y = ((5x^2 +2x−1)/(x−2))


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 123407 by bemath last updated on 25/Nov/20

$F i n d t h e a l l a s y m t o t e s o f t h e$ $c u r v e y = \frac{5 x^{2} + 2 x - 1}{x - 2}$
Commented by EVIMENEBASSEY last updated on 25/Nov/20

$F i n d t h e a l l a s y m t o t e s o f t h e$ $c u r v e y = \frac{5 x^{2} + 2 x - 1}{x - 2}$ $solution$ $y (x - 2) = {5 x}^{2} + 2 x - 1$ $horizontal asymtote = 2$ $no horizontal asymtote .$ $but letting y = mx + c$ $(mx + c) (x - 2) = {5 x}^{2} + 2 x - 1$ ${mx}^{2} - mx + xc - 2 c = {5 x}^{2} + 2 x - 1$ $c = m = 0$ $caddet$
Commented by liberty last updated on 25/Nov/20
$(i) vertical asymtotes : lim_(x→2) y = lim_(x→2) ((5x^2 +2x−1)/(x−2)) = ∞ we get vertical asymtotes x=2 (ii) horizontal asymtotes : lim_(x→±∞) y =lim_(x→±∞) ((5x^2 +2x−1)/(x−2)) clearly the limit does not exist so the curve no horizontal asymtotes (iii) we want find inclined asymtotes y = kx + b where { ((k=lim_(x→∞) (y/x)=lim_(x→∞) ((5x^2 +2x−1)/(x^2 −2x))=5)),((b=lim_(x→∞) (y−kx)=lim_(x→∞) (((5x^2 +2x−1)/(x−2))−5x))) :} ⇒b = lim_(x→∞) (((5x^2 +2x−1−5x^2 +10x)/(x−2))) ⇒b = lim_(x→∞) (((12x−1)/(x−2)))=12 thus straight line y = 5x+12 is inclined asymtotes$
$(i) v e r t i c a l a s y m t o t e s : \lim_{x \to 2} y = \lim_{x \to 2} \frac{5 x^{2} + 2 x - 1}{x - 2} = \infty$ $w e g e t v e r t i c a l a s y m t o t e s x = 2$ $(i i) h o r i z o n t a l a s y m t o t e s : \lim_{x \to \pm \infty} y = \lim_{x \to \pm \infty} \frac{5 x^{2} + 2 x - 1}{x - 2}$ $c l e a r l y t h e l i m i t d o e s n o t e x i s t s o$ $t h e c u r v e n o h o r i z o n t a l a s y m t o t e s$ $(i i i) w e w a n t f i n d i n c l i n e d a s y m t o t e s$ $y = k x + b w h e r e {\begin{cases} k = \lim_{x \to \infty} \frac{y}{x} = \lim_{x \to \infty} \frac{5 x^{2} + 2 x - 1}{x^{2} - 2 x} = 5 \\ b = \lim_{x \to \infty} (y - k x) = \lim_{x \to \infty} (\frac{5 x^{2} + 2 x - 1}{x - 2} - 5 x) \end{cases}$ $\Rightarrow b = \lim_{x \to \infty} (\frac{5 x^{2} + 2 x - 1 - 5 x^{2} + 10 x}{x - 2})$ $\Rightarrow b = \lim_{x \to \infty} (\frac{12 x - 1}{x - 2}) = 12$ $t h u s s t r a i g h t l i n e y = 5 x + 12 i s$ $i n c l i n e d a s y m t o t e s$
Commented by MJS_new last updated on 25/Nov/20
$y=((p(x))/(q(x))) with p, q are polynomes and degree (p) ≥degree (q) you can always find existing asymptotic lines or curves by dividing the fraction y=((7x+3)/(x−5))=7+((38)/(x−5)) ⇒ asymptotes { ((x=5)),((y=7)) :} y=((7x^2 +3)/(x−5))=7x+35+((178)/(x−5)) ⇒ asymptotes { ((x=5)),((y=7x+35)) :} y=((7x^3 +3)/(x^2 −5))=7x+((35x+3)/(x^2 −5)) ⇒ asymptotes { ((x=±(√5))),((y=7x)) :} asymptotic curves: y=((7x^3 +3)/(x−5))=7x^2 +35x+175+((878)/(x−5)) ⇒ asymptotes { ((x=5)),((y=7x^2 +35x+175)) :} y=((7x^4 +3)/(x^2 −5))=7x^2 +35+((178)/(x^2 −5)) ⇒ asymptotes { ((x=±(√5))),((y=7x^2 +35)) :} ...$
$y = \frac{p (x)}{q (x)} with p, q are polynomes and degree (p) ⩾ degree (q)$ $you can always find existing asymptotic lines$ $or curves by dividing the fraction$ $y = \frac{7 x + 3}{x - 5} = 7 + \frac{38}{x - 5} \Rightarrow asymptotes {\begin{cases} x = 5 \\ y = 7 \end{cases}$ $y = \frac{7 x^{2} + 3}{x - 5} = 7 x + 35 + \frac{178}{x - 5} \Rightarrow asymptotes {\begin{cases} x = 5 \\ y = 7 x + 35 \end{cases}$ $y = \frac{7 x^{3} + 3}{x^{2} - 5} = 7 x + \frac{35 x + 3}{x^{2} - 5} \Rightarrow asymptotes {\begin{cases} x = \pm \sqrt{5} \\ y = 7 x \end{cases}$ $asymptotic curves :$ $y = \frac{7 x^{3} + 3}{x - 5} = 7 x^{2} + 35 x + 175 + \frac{878}{x - 5} \Rightarrow asymptotes {\begin{cases} x = 5 \\ y = 7 x^{2} + 35 x + 175 \end{cases}$ $y = \frac{7 x^{4} + 3}{x^{2} - 5} = 7 x^{2} + 35 + \frac{178}{x^{2} - 5} \Rightarrow asymptotes {\begin{cases} x = \pm \sqrt{5} \\ y = 7 x^{2} + 35 \end{cases}$ $. . .$
Commented by bemath last updated on 25/Nov/20

Commented by bemath last updated on 25/Nov/20

Commented by MJS_new last updated on 25/Nov/20

$for y = \frac{7 x^{4} + 3}{x^{2} - 5} and y = 7 x^{2} + 35 you will have to$ $set up a different window . try$ $- 4 ⩽ x ⩽ 4 and - 100 ⩽ y ⩽ 300$
Commented by bemath last updated on 25/Nov/20

$y e s s i r .$
	Answered by MJS_new last updated on 25/Nov/20

	$vertical asymptote x = 2$ $y = 5 x + 12 + \frac{23}{x - 2} \Rightarrow asymptote y = 5 x + 12$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum