- BloomLocke367

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

- schrodinger

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- Nnesha

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

- Nnesha

XD

- anonymous

can u help me @Nnesha

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## More answers

- BloomLocke367

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.

- Nnesha

that's right !!

- BloomLocke367

wow, okay. that was easy.

- BloomLocke367

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

- Nnesha

you got it!

- BloomLocke367

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)

- Nnesha

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

- BloomLocke367

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

- BloomLocke367

thanks for your help btw

- Nnesha

is there a denominator ?

- BloomLocke367

no? just 1. lol

- Nnesha

there is just one right
so no vertical asy

- BloomLocke367

okay

- Nnesha

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

- BloomLocke367

NO HORIZONTAL ASYMPTOTE.

- BloomLocke367

sorry, I had that in caps in my notes XD

- Nnesha

yes right

- BloomLocke367

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.

- BloomLocke367

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.

- Nnesha

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

- Nnesha

yes you're right!

- BloomLocke367

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

- Nnesha

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

- BloomLocke367

okay thank you

- BloomLocke367

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.

- Nnesha

|dw:1441231766886:dw|
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

- Nnesha

aw np :=)

- BloomLocke367

I have to go eat dinner, thank you!

- Nnesha

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

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