BloomLocke367
  • BloomLocke367
How do you tell from looking at an equation whether there is a vertical or horizontal asymptote?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Nnesha
  • 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
  • Nnesha
XD
anonymous
  • anonymous
can u help me @Nnesha

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

BloomLocke367
  • 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
  • Nnesha
that's right !!
BloomLocke367
  • BloomLocke367
wow, okay. that was easy.
BloomLocke367
  • BloomLocke367
so for \(q(x)=\large \frac{x-1}{x}\) vertical: x=0 and horizontal:y=1?
Nnesha
  • Nnesha
you got it!
BloomLocke367
  • 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
  • Nnesha
that's right! -x+3= -3 so x= 3
BloomLocke367
  • BloomLocke367
This one I don't know at all :O \(q(x)=1.5^x\)
BloomLocke367
  • BloomLocke367
thanks for your help btw
Nnesha
  • Nnesha
is there a denominator ?
BloomLocke367
  • BloomLocke367
no? just 1. lol
Nnesha
  • Nnesha
there is just one right so no vertical asy
BloomLocke367
  • BloomLocke367
okay
Nnesha
  • Nnesha
N>D ( highest degree) numerator is greater than denominator so horizontal asy would bE ?
BloomLocke367
  • BloomLocke367
NO HORIZONTAL ASYMPTOTE.
BloomLocke367
  • BloomLocke367
sorry, I had that in caps in my notes XD
Nnesha
  • Nnesha
yes right
BloomLocke367
  • 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
  • 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
  • Nnesha
yes h- asy y=0 right but when you solve for x you will get negative one x^2=-1
Nnesha
  • Nnesha
yes you're right!
BloomLocke367
  • BloomLocke367
can you have an imaginary asymptote? or is the answer none?
Nnesha
  • Nnesha
you will get x= i which is imaginary i would say no vertical asy
BloomLocke367
  • BloomLocke367
okay thank you
BloomLocke367
  • 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
  • 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
  • Nnesha
aw np :=)
BloomLocke367
  • BloomLocke367
I have to go eat dinner, thank you!
Nnesha
  • Nnesha
np :=) have a wonderful evening o^_^o

Looking for something else?

Not the answer you are looking for? Search for more explanations.