Question

Give an example of a rational function that has no horizontal asymptote and a vertical asymptote at
x = 1.

Answer 1

That means that \(x = 1\) is an excluded value.

Answer 2

im not really sure what u mean

Answer 3

That means that we won't have \(x = 1 \) in the domain.\[{x^2 \over x - 2} \]Has a vertical asymptote at \(x = 2\).

Answer 4

thats it?

Answer 5

Yep, just that you have to find it for \(x = 1\).

Answer 6

so i can just put x^2/x-2

Answer 7

Nope.

Answer 8

Hint: You can't input 1 in\[x^2\over x - 1 \]

Answer 9

im lost at this point

Answer 10

Parth was giving you an example of the concept, and then hoping that you would then apply it to your particular problem.

Answer 11

right but I'm not really understanding how to do the question i need help finding the example of what they are asking for

Answer 12

Do you know what the 'excluded value' means?

Answer 13

Excluded values are simply that: values that are excluded, or left out. These are values that will make the denominator of a rational expression equal to 0.

Answer 14

Exactly. The question merely means that the excluded value of \(x\) is \(1\).

Answer 15

Well there you go. Make an equation such that the denominator is equal to 0 when x=1, and then just make sure that it doesn't have a horizontal asymptote.

Answer 16

you could look at it in this way:\[f(x)=\frac{g(x)}{h(x)}\]find a function \(h(x)\) such that \(h(1)=0\)

Answer 17

4x^4y^3
______
-8x^2y

Answer 18

Woo! That's a great explanation, but it's a little complicating for a beginner.

Answer 19

KISS principle: http://en.wikipedia.org/wiki/KISS_principle

Answer 20

keep it simple :)

Answer 21

You can even make it something like \[x\over x - 3 \]If 3 is the horizontal asymptote :P

Answer 22

@asnaseer That KISS Principle link was for @babydoll332 haha
She posted a little too complicated expression.

Answer 23

yes I /guessed/ it was :)

Answer 24

@babydoll332 maybe a different approach might help.

if:\[y=x-a\]where a is some constant, then what value of x will make y zero?

Answer 25

Just wanted to make sure that you didn't take that. All swell! :D

Answer 26

so is my expression right or wrong or is it to complex for u to explain that @ParthKohli

Answer 27

your expression is wrong

Answer 28

The point is that there are flaws in your expression - and also you need to keep your answers simple! :)

Answer 29

and far too complex than required

Answer 30

@asnaseer Congratulations for becoming a champ!

Answer 31

?

Answer 32

ok then give me an easier example so i can make it the way i want after

Answer 33

You're a champion at problem solving haha

Answer 34

I did give one: If x = 3 is a vertical asymptote, then we can make the function look like\[f(x) = {x \over x - 3} \]

Answer 35

thats the it thats the whole thing way the question is asking for

Answer 36

Because then, plugging 3 into \(x\) will make the denominator 0.
You did post how the denominator is 0 in a vertical asymptote, didn't you?

Answer 37

You have to make an expression where plugging 1 in \(x\) makes the denominator 0. All right?

Answer 38

Parth - you also need to ensure it has no horizontal symptotes

Answer 39

u are confusing me omg !!!

Answer 40

@babydoll332 sorry if my comments confused you further. maybe I will let Parth explain this to you.

Answer 41

I was going to leave it to you >.>

Answer 42

nooo i want @asnaseer

Answer 43

Maybe it's my words that confused her.

Answer 44

I don't mind trying to explain it as long as my friend Parth agrees. :)

Answer 45

Continue, I am watching you guys\[\Huge \ddot \smile \]

Answer 46

this is my example

x+1 /x(x+4)

Answer 47

Keep in mind: Does plugging in 1 make the denominator 0?

Answer 48

ok, now if we take your example, we see that at x=1 we get:\[\frac{1+1}{1(1+4)}=\frac{2}{5}\ne\infty\]

Answer 49

you need to find an equation such that when you set x=1, the denominator will become zero.

Answer 50

so lets first look at a simple function.

y = x - a

what value of x makes y zero?

Answer 51

(x-1)(x+2)

Answer 52

yes that would work as the denominator

Answer 53

so now you know that you have something of the form:\[y=\frac{g(x)}{(x-1)(x+2)}\]where g(x) is some other function of x.
the only problem with the example you have selected is that it makes the denominator equal to zero at x=1 AND at x=-2

Answer 54

so this has TWO vertical asymptotes - which is not what was asked for.

Can you think of a simpler example?

Answer 55

(x-1)(x-2)?

Answer 56

that still has TWO vertical asymptotes - one at x=1 and the other at x=2

Answer 57

think even simpler

Answer 58

so (x+1)(x+2)?

Answer 59

that will have asymptotes at x=-1 and x=-2
but you were asked to find one that has an asymptote at x=1

it looks like you are trying to get some quadratic equation to fit this question. remember it does not have to be quadratic. think of a polynomial of degree 1.

Answer 60

does that make sense?

Answer 61

a polynomial of degree 1 would be of the form:\[ax+b\]where a and b are constants.

Answer 62

(x+1)/(x^2-4)

Answer 63

Remember that x + 1 should be the denominator.

Answer 64

You must get the denominator as 0, right?

Answer 65

i guess i don't know thats why i am asking because i don't get the question nor what the example would be

Answer 66

the example you just gave doesn't work as it makes the denominator zero when x=2 or x=-2

Answer 67

Reread the explanations that asnaseer and parth have both given you. Then reread the section of your textbook that talks about this subject. Then think about the subject for ten minutes. Then come back, and if you still don't get it, ask specific questions about specific parts that you do not understand.

Answer 68

let me try one more approach first..

Answer 69

i just want an example thats all I'm asking for so then i can go and actually apply it and understand what both of you are trying to explain

Answer 70

1) do you agree that you will get a vertical asymptote if you have a rational function where the denominator can become zero?

Answer 71

They've given you roughly a dozen examples by now.

Answer 72

ok yea

Answer 73

ok

Answer 74

good, now, in order to get a vertical asymptote at x=1, you need to find a function in the denominator that will become zero when x=1.

agreed?

Answer 75

agreeed

Answer 76

great - we are ALMOST there now...

Answer 77

what is the SIMPLEST function you can think of that will become zero when x=1?

Answer 78

x=o

Answer 79

that is just stating x=0

you need to create a function that, when you plug x=1 into it, will give you a value of zero.

Answer 80

e.g. if the function were y = x - 3
then this gives a value of zero when x=3

Answer 81

because at x=3 you get y = 3 - 3 = 0

Answer 82

ok so y=x-1

Answer 83

perfect!

Answer 84

so we have now defined the function for the denominator - it is (x-1)

Answer 85

the next step is to find a function for the numerator such that we do NOT get any horizontal asymptotes

Answer 86

do you know what properties of a function will give a horizontal asymptote?

Answer 87

not really

Answer 88

ok, let me try to explain it...

Answer 89

we know your final function has to be of the form:\[y=\frac{g(x)}{x-1}\]where g(x) is the function we need to find for the numerator.

now, if y tends to some fixed value as x tends to plus/minus infinity, then this means that y has a horizontal asymptote

Answer 90

I think the easier explanation for someone at this level might have to use graphing =/

Answer 91

e.g. lets say you had:\[y=\frac{x}{x-1}\]then can you see that as x tends to infinity then y tends to 1?

Answer 92

good point nbouscal - let me draw a graph...

Answer 93

I'm good i don't need to see a graph

Answer 94

How are you going to solve it, then?

Answer 95

are you sure babydoll332?

Answer 96

ya i think I'm good

Answer 97

Ok, what is the solution then??