Question

Consider the function f(x)= x^2 -16 / x-2 Determine all asymptotes of this function including horizontal, vertical, and oblique (slant).

Answer 1

Do you know how to take limits?

for Vertical asymptotes you take limit of numbers not found in the domain

for Horizontal asymptotes you take the limit at + and - infinity

Answer 2

so for this case take

lim
x->2^(+)
and
lim
x->2^(-)

if you get infinity then an asymptote exists

Answer 3

I've never taken limits.

Answer 4

for horizontal asymptotes

lim
x-> infinity

lim
x-> -infinity

if you get infinity out you do not have an asymptote, if you get a number out an asymptote exists at that number

Answer 5

why you even doin homework over the summer

Answer 6

lim x^2 -16 / x-2
x->2^(+)

so we insert 2^(+) into the function

(2^(+))^(2)-16/x-2^(+)
=
-12/0^(+)
how many times does a infinity small number go into -12, infinite times thus
= infinity

what we mean by 2^(+) is a number a little greater than 2 because we can compute a value in the function for something infinity close to 2, but not for 2 because you cannot divide by zero

Answer 7

I'm taking extra classes to boost my gpa.

Answer 8

Since f(x) = [(x^2 + 4)(x + 2)(x - 2)]/(x - 2) = (x^2 + 4)(x + 2) for x <> 2. So the domain is all real numbers excepts x = 2. The graph has a whole when x = 2 but has no asymptote.

Answer 9

by taking the limit of 2^(+) we see that it is -negative infinity thus we know that an asymptote exists at 2, and that to the right of 2 the function is decreasing towards infinity

Answer 10

Mr. To the graph has asymptotes
https://www.wolframalpha.com/input/?i=%28x^2+-16+%29%2F%28x-2%29

Answer 11

furthermore I just showed it contained a vertical asymptote

Answer 12

Watch this guys videos

https://www.youtube.com/watch?v=HeqfhnKncjc

Answer 13

he can explain this stuff better than I

Answer 14

https://www.youtube.com/watch?v=c-yK2hUnSB0

Answer 15

My bad. I saw x^4 - 16/x-2

Answer 16

https://www.youtube.com/watch?v=_qEOZNPce60

Answer 17

Oh dear. Thank you all for trying to help!

Answer 18

Mr. To even if it was (x^(4) - 16)/(x-2) it would still contain an asymptote

Answer 19

Tiffany watch those videos if you wish to understand this material better its what I did to learn this stuff this guy is a very good teacher

Answer 20

Thank you

Answer 21

@Mr._To
not to discourage you but if you do not have a very intimate understanding of the material in a question you should with hold trying to solve it until you are certain you are correct, as you will just confuse people and drown out people who know what they are talking about

Answer 22

vertical asymptote: x=2
horizontal asymptote=slant : y= x+2

Answer 23

I mean posting an answer

Answer 24

Michelle, How did you do that?

Answer 25

Watch these three videos they will explain how to find asymptotes and they are under short
https://www.youtube.com/watch?v=c-yK2hUnSB0
https://www.youtube.com/watch?v=HeqfhnKncjc
https://www.youtube.com/watch?v=_qEOZNPce60

Answer 26

very*

Answer 27

@ Australopithecus
I don't want to interrupt your "teaching". Buy since you brought it up, I would like to say a few words. I did apologize that I misread as (x^4 - 16)(x-2). Therefore, I factored the numerator into (x^2 + 4)(x^2 - 4) = (X^2 + 4)(x + 2)(x - 2) then simplify with the denominator to get (X^2 +4)(x+2). In this case, the graph has no asymptote but only a whole at x = 2.
Again, I apologize for misreading the problem, but...
Learning math, we don't jump to a conclusion without sufficient evidences and proofs. We don't know each other. How do you know I don't understand the material?
Any way, I apologize.

Answer 28

you just have to follow a few absolute rules... to find a vertical asymptote in a rational function, take JUST the denominator and make it = 0 then youll find the vertical asymptote(s). horizontal however becomes more tricky. the rules are: RULE 1: if the highest exponent of each equation in the numerator and denominator are equal then the horizontal asymptote will bethe coefficient of the highest exponent in the numerator divided by the coefficient of the highest exponent in the denominator. eg: 2x^2 / 3x^2...... exponents both equal 2 therefore the horizonal asymptote will = 2/3. RULE 2: if the highest exponent of the denominator is larger than the higherst exponent in the numerator then the horizonatl asymptote will equal y=0. Rule 3: (this is the one i used in your question...) if the exponent in the numerator is higher than in the denominator, you will have to find a slant function by means of long devision. you will have to long divide in your example x^2 -16 by x-2. you will then get the slant equation of y=x+2 with a remainder of -12 . dont worrie about the remainder untill you have to describe the limitations of your function.

Answer 29

send me a message if you dont understand and ill try clarify for you. good luck