## BloomLocke367 one year ago How do you tell from looking at an equation whether there is a vertical or horizontal asymptote?

1. 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

2. Nnesha

XD

3. anonymous

can u help me @Nnesha

4. 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.

5. Nnesha

that's right !!

6. BloomLocke367

wow, okay. that was easy.

7. BloomLocke367

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

8. Nnesha

you got it!

9. 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)

10. Nnesha

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

11. BloomLocke367

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

12. BloomLocke367

13. Nnesha

is there a denominator ?

14. BloomLocke367

no? just 1. lol

15. Nnesha

there is just one right so no vertical asy

16. BloomLocke367

okay

17. Nnesha

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

18. BloomLocke367

NO HORIZONTAL ASYMPTOTE.

19. BloomLocke367

sorry, I had that in caps in my notes XD

20. Nnesha

yes right

21. 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.

22. 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.

23. Nnesha

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

24. Nnesha

yes you're right!

25. BloomLocke367

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

26. Nnesha

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

27. BloomLocke367

okay thank you

28. 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.

29. 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

30. Nnesha

aw np :=)

31. BloomLocke367

I have to go eat dinner, thank you!

32. Nnesha

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