Question

How do you tell from looking at an equation whether there is a vertical or horizontal asymptote?

Answer 1

for \(\color{green}{\rm Vertical~ asy.}\)
set the denominator equal to zero and then solve for the variable.
for\(\color{green}{\rm Horizontal ~asy.}\)
focus on highest degrees
~if the highest degree of the numerator is greater than the denominator then `No horizontal asy.` \[\color{reD}{\rm N}>\color{blue}{\rm D}\] example
\[\large\rm \frac{ 7x^\color{ReD}{3} +1}{ 4x^\color{blue}{2}+3 }\]
~if the highest degree of the denominator is greater than the highest degree of the numerator then `y=0` would be horizontal asy.
\[\rm \color{reD}{N}<\color{blue}{\rm D}\]
example:\[\large\rm \frac{ 7x^\color{red}{2}+1 }{ 4x^\color{blue}{3}+3 }\]
~if both degrees are the same then divide the leading coefficient of the numerator by the leading coefficient of the denominator
\[\rm \color{red}{N}=\color{blue}{D}\]
\[\large\rm \frac{ 8x^\color{reD}{3}+1 }{ 4x^\color{blue}{3}+3 }\] \[\rm \frac{ 8x^3 }{ 4x^3 } =2\] horizontal asy. =2

Answer 2

XD

Answer 3

can u help me @Nnesha

Answer 4

okay.. so \(f(x)=\frac{x}{x-1}\) the vertical asymptote is x=1 and the horizontal is y=1, right? because the highest degrees are the same and are both 1. The leading coefficients are also both 1 and 1/1=1 for the vertical asymptote 1-1=0.

Answer 5

that's right !!

Answer 6

wow, okay. that was easy.

Answer 7

so for \(q(x)=\large \frac{x-1}{x}\) vertical: x=0 and horizontal:y=1?

Answer 8

you got it!

Answer 9

okay. this one is a little different but I'm PRETTY sure I got it. \(g(x)=\large\frac{x+2}{3-x}\)...so v. tote: x=3 and h. tote: y=-1? (I shortened it lol)

Answer 10

that's right!
-x+3= -3
so x= 3

Answer 11

This one I don't know at all :O \(q(x)=1.5^x\)

Answer 12

thanks for your help btw

Answer 13

is there a denominator ?

Answer 14

no? just 1. lol

Answer 15

there is just one right
so no vertical asy

Answer 16

okay

Answer 17

N>D ( highest degree)
numerator is greater than denominator so horizontal asy would bE ?

Answer 18

NO HORIZONTAL ASYMPTOTE.

Answer 19

sorry, I had that in caps in my notes XD

Answer 20

yes right

Answer 21

okay. now for this one, \(p(x)=\large\frac{4}{x^2+1}\) the horizontal asymptote is y=0 because it's "heavy on the bottom" (there is a higher degree on the bottom) but I'm not sure about the vertical asymptote.

Answer 22

because it would be \(x^2=-1\) which would be i for the answer which is an imaginary number. I don't think you can have x=i, but I may be wrong.

Answer 23

yes h- asy y=0 right
but when you solve for x you will get negative one
x^2=-1

Answer 24

yes you're right!

Answer 25

can you have an imaginary asymptote? or is the answer none?

Answer 26

you will get x= i which is imaginary
i would say no vertical asy

Answer 27

okay thank you

Answer 28

I think I'm good. I did three more but it wouldn't be fair for me to have ALL of my homework checked XD thank you.

Answer 29

graph (from purplemath.com)
in this graph red lines represent vertical asy
and we can't see imaginary stuff so i would say not vertical asy

Answer 30

aw np :=)

Answer 31

I have to go eat dinner, thank you!

Answer 32

np :=) have a wonderful evening o^_^o