Question

y=2x-3/x+4 What is the equation of this functions vertical asymptote?

Answer 1

You find vertical asymptotes by finding the values of x that make the function undefined. So for this, the function will be undefined whenever the denominator is 0. SO that being said, we need to take the denominator (x+4), set it equal to 0 and then solve for x. Once you do that, your value of x will be where you have a vertical asymptote.

Answer 2

your fast

Answer 3

x=-4

Answer 4

Lol, Ididn't even notice this was your question x_x I just saw it and came here, haha. And yes, x= -4 is whereyour asymptote is :3

Answer 5

funny! does this function have any additional asymptote like horizontal or oblique?

Answer 6

horizontal... y=2

Answer 7

is it 3/2

Answer 8

Horizontal yes. You have a horizontal asymptote when the degree of the numerator and denominator are the same. In this case, the highest power of x is 1 in the top and bottom, so your horizontal asymptote is dividing the coefficients of the highest power of x. As pg just posted, it would be y =2, since you have 2x divided by x.

Answer 9

ow thanks @pgpilot326

Answer 10

@Psymon when x has the same power right.

Answer 11

yes, but you'll also have a horizontal asymptote at y=0 when the degree of the numerator is less than the degree of the denominator... just like y = 1/x

Answer 12

Well, when the highest power in the numerator and the denominator are the same you have a horizontal asymptote that you can find by dividing thecoefficients of those powers. And yeah, I always got someone watching over me to keep correcting my mistakes. You would also have one at y = 0 if the denominator is of a higher degree than the numeraotr.

Answer 13

@pgpilot326 how did u get y=0

Answer 14

Sorry, I'll back off, pg has this better than I do x_x

Answer 15

as x gets bigger , y = 1/x gets closer and closer to 0.

Answer 16

no... you've got @Psymon

Answer 17

i'll split.

Answer 18

no @Psymon your helpful to you give good explanations too.

Answer 19

Lol, I just lack confidence and figure if anyone will be wrong about something it'll be me xD

Answer 20

@Psymon and @pgpilot326 none of you guys have been wrong and you guys can team up.

Answer 21

Of course we can, I'm just paranoid xD

Answer 22

no... trust me. you're doing great. some people here expect or want perfection but it's difficult to be that.
I got pounced on for making mistakes but unfortunately it's part of the process. however the pouncing part shouldn't be... after all, I would hope everyone is here to learn.

Answer 23

@Psymon your being really silly there is no reason to be paranoid.

Answer 24

so please, @Psymon continue...

Answer 25

Lol, you don't need to explain that to me :P I'm just that way, I know I'm irrational in my thinking, haha.

Answer 26

ya both of you guys continue.

Answer 27

i just don't want to make anyone feel like others made me feel earlier today.

Answer 28

Well.....we finished up the horizontal asymptotes. As for the last kind of asymptote, oblique asymptotes, this would occur when the numerator has a degree exactly one higher than the denominator. If this is the case, then you find the equation for the oblique asymptote via long division. You divide the denominator into the numerator and ignore any remainder. What you come out with is your equation for the oblique asymptote :3

And sucks to here, pg D: I've left here for the day not always feeling too great myself x_x

Answer 29

Dont worry about it even that happen to me and all i care about is if they got the help they needed.

Answer 30

it's okay... good job on finishing up. Couldn't have said it any better! Keep up the good work and did you follow all of that @Rosa34 ?

Answer 31

I'm just more sensitive than most, just something I have to deal with xD But yeah, I know the situation didn't come up, but values of x that make the denominator 0 can be holes instead of asymptotes at times, too o.o

Answer 32

I really think your smart @Psymon and @pgpilot326 just take the positive with you at the end of the day.

Answer 33

thanks @Rosa34 , you too!

Answer 34

your welcome

Answer 35

^_^

Answer 36

That's right @Psymon! I think that happens when the factor(s) cancel.

Answer 37

sorry got lost.

Answer 38

you get the missing hole vs. the asymptote

Answer 39

Right. Feel like it's important to mention that.

So yeah, you can sometimes have holes in your graph instead of asymptotes @Rosa34

Answer 40

Here's an example:

Answer 41

suppose y = (x^2-1)/(x-1) this is just x+1 with a hole at x = 1.

Answer 42

ok i think i understand

Answer 43

sorry... didn't see cause i was typing. @Psymon probably has a better example and pic(?)

Answer 44

\[\frac{ x ^{2}+5x + 6 }{ x+2 }\] Now I would want to factor the top first before I start making any assumptions because if the denominator cancels, I have a hole instead of an asymptote

\[\frac{ (x+3)(x+2) }{ (x+2) }\]
Now -2 would make the denominator 0, so we would be undefined, but since that factor cancels out with the (x+2) factor on top, the -2 is a hole instead of asymptote. So basically you need to check and see what in the bottom cancels. If it cancels out, it's a hole, if it does not cancel out, it is a horizontal asymptote.

Answer 45

very well said!!!

Answer 46

*vertical

Answer 47

yes i got it thanks for both examples

Answer 48

Alrighty, awesome :3

Answer 49

ok to sum it up yes there is a horizontal asymptote at y=2 and y=0.

Answer 50

No, not both.

Answer 51

only at y=2

Answer 52

Correct :3

Answer 53

and there is no oblique asymptotes

Answer 54

Correct. THat would only occur if the degree of the numerator was exactly one higher than the denominator.

Answer 55

well, no. it's the line itself.

Answer 56

in response to @Rosa34

Answer 57

got it and to get y=2 you divide numorator and dinaminator which is 2/1

Answer 58

Well, you divide the coefficients of the highest powers.

Answer 59

yeah as x goes to + or - infinity.

Answer 60

like @Psymon said

Answer 61

ok i understand

Answer 62

so if i were to graph my horizontal and vertical asymptotes would it be two lines or the same line?

Answer 63

Well, you would draw a dashed line straight up and down through x = -4 and a dashed line straight left and right at y = 2.

Answer 64

thank u very much @Psymon and @pgpilot326 for all the help.

Answer 65

Mhm mhm ^_^

Answer 66

you're welcome

Answer 67

so @Psymon i still didnt finish the problem we were working on last nigh. I really dont understand all of it.

Answer 68

I don't recall what it was.

Answer 69

it was about the vertices

Answer 70

Ah. I just forget which conic section it was. The hyperbola, right?

Answer 71

yes u are correct again.

Answer 72

Alrighty. Well, I'm not positive what exactly the problem was, but I can try and go over the rules for hyperbolas.

Answer 73

ok so we got c=5 because a= 9, b=16, you use a^2+b^2=c^2

Answer 74

Ah, so this was
\[\frac{ y ^{2} }{ 9 }-\frac{ x ^{2} }{ 16 }=1\]
And yes, hyperbola gives you foci by a^2 + b^2 = c^2

Answer 75

\[(y+3)^2/9- (x-2)^2/16=1\]

Answer 76

ok what about vertices?

Answer 77

Oh, okay, so the center was elsewhere xD So yeah,center at 2, -3. Since this has y^2 as the positive fraction, the verticies and foci are up and down from the center. The verticies are up and down a units from the center. Now, a^2 is always the number in the denominator of the positive fraction, which means a^2 is 9 and a is 3. So up 3 units and down 3 units from the center our are verticies.