## anonymous one year ago Will METAL!! For the given function, find the vertical and horizontal asymptote(s) (if there are any). f(x) = the quantity two x squared plus one divided by the quantity x squared minus four None x = 2, y = 2y = 0 x = 2, x = -2, y = 2 x = 2, y = 2, y = 1

1. anonymous

|dw:1443803129894:dw|

2. 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

3. Nnesha

looks vry long but it's not :D

4. anonymous

so x=2

5. Nnesha

remember when we take square root we should get plus/minus answer $\huge\rm \sqrt{x^2} = \pm x$

6. anonymous

+-2?

7. Nnesha

right

8. anonymous

So it would be my third answer choice because that's the only one with a negative number.

9. anonymous

right?

10. Nnesha

lol right what about horizontal asy ?

11. anonymous

the horizontal asy is 2 because both degs on the top and bottom are even. Right?

12. Nnesha

highest degree of both (Denominator and numberatr) are the same so just divide the leading coefficient |dw:1443803913846:dw|