anonymous
  • anonymous
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
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1443803129894:dw|
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
looks vry long but it's not :D

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anonymous
  • anonymous
so x=2
Nnesha
  • Nnesha
remember when we take square root we should get plus/minus answer \[\huge\rm \sqrt{x^2} = \pm x\]
anonymous
  • anonymous
+-2?
Nnesha
  • Nnesha
right
anonymous
  • anonymous
So it would be my third answer choice because that's the only one with a negative number.
anonymous
  • anonymous
right?
Nnesha
  • Nnesha
lol right what about horizontal asy ?
anonymous
  • anonymous
the horizontal asy is 2 because both degs on the top and bottom are even. Right?
Nnesha
  • Nnesha
highest degree of both (Denominator and numberatr) are the same so just divide the leading coefficient |dw:1443803913846:dw|

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