)Question) Find the horizontal and vertical Asymptotes of f(x)=xx2+4 (1) y=-1, x=-1(2)y=-1, x=1(3) No horizontal asymptote, x=-1(4) No vertical asymptote, - Maths - Limits and Derivatives

Question

) Q u e s t i o n )   F i n d   t h e   h o r i z o
n t a l   a n d   v e r t i c a l   A s y m p t o t
e s   o f   f ( x ) = x x 2 + 4 ( 1 )   y = - 1 ,
  x = - 1 ( 2 ) y = - 1 ,   x = 1 ( 3 )   N o  
h o r i z o n t a l   a s y m p t o t e ,   x = - 1 ( 4 )
  N o   v e r t i c a l   a s y m p t o t e ,  
x = - 1

Shivam Mishra · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

y=xx2+4limx→-∞ xx2+4=limx→-∞ xx1+4x2=limx→-∞ -11+4x2 Aslim x→-∞ , hence xx=-1=-11+4∞=-11+0y=-1As fx is defined for all real x hence there will be no vertical asymptote
Option4 is correct

Hope this information will clear your doubts about this topic

If you have any doubts just ask here on the ask and answer forum
and our experts will try to help you out as soon as possible
Regards

) Q u e s t i o n ) F i n d t h e h o r i z o n t a l a n d v e r t i c a l A s y m p t o t e s o f f ( x ) = x x 2 + 4 . ( 1 ) y = - 1 , x = - 1 ( 2 ) y = - 1 , x = 1 ( 3 ) N o h o r i z o n t a l a s y m p t o t e , x = - 1 ( 4 ) N o v e r t i c a l a s y m p t o t e , x = - 1

) $Q u e s t i o n) F i n d t h e h o r i z o n t a l a n d v e r t i c a l A s y m p t o t e s o f f (x) = \frac{x}{\sqrt{x^{2} + 4}} . (1) y = - 1, x = - 1 (2) y = - 1, x = 1 (3) N o h o r i z o n t a l a s y m p t o t e, x = - 1 (4) N o v e r t i c a l a s y m p t o t e, x = - 1$