Sketch graphs showing vertical & horizontal asymptotes and relative extrema

Click For Summary
The discussion focuses on sketching the graph of the function f(x) = (x^2-1)/(x^2-4), specifically identifying vertical and horizontal asymptotes and relative extrema. The vertical asymptotes are correctly identified at x = ±2, but there is confusion regarding the behavior of the function near x = -2. The limits as x approaches -2 from the right and left are confirmed to be +infinity and -infinity, respectively. The participant expresses uncertainty about their graphing calculator's output, which contradicts these limits. Clarification is provided that the limits are accurate, suggesting the participant may need to revisit their graphing approach.
Glissando
Messages
34
Reaction score
0

Homework Statement


Sketch the graphs of the following function showing vertical and horizontal asymptotes and relative extrema:

f(x) = (x2-1)/(x2-4)


Homework Equations


Limits, zeroes, derivatives


The Attempt at a Solution


I know that I have the majority of the answers right, the only problem I'm having is with part of the vertical asymptote.

VA = positive and negative 2

The problem is with -2:

lim (-22-1)/(-2+)2-4)
x->-2+

= (4-1)/(4+-4) = +infinity

lim ((-2)2-1)/((-2-)2-4)
x->-2-

= (4-1)/(4--4) = negative infinity

But my graphing calculator shows the exact opposite ):!

Any help is appreciated! Thank you <3
 
Physics news on Phys.org
The limit as x->(-2)+ is +infinity. The limit as x->(-2)- is -infinity. I'm not sure what your graphing calculator's problem is.
 
Dick said:
The limit as x->(-2)+ is +infinity. The limit as x->(-2)- is -infinity. I'm not sure what your graphing calculator's problem is.

...I guess I'm sketching the graph wrong then...Thank you for your help (:!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K