anonymous
  • anonymous
how do i find the vertical asymptote of the function f(x)=4/(x+5)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
you first cross out like terms from top to bottom, then what ever value makes the bottom a zero is your VA
amistre64
  • amistre64
in this case....x =-5
anonymous
  • anonymous
so you just make x+5=0

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anonymous
  • anonymous
then solve for x?
anonymous
  • anonymous
Try finding the lim as x goes to 0
amistre64
  • amistre64
that is correct, you want to know what vale of x can never be used
amistre64
  • amistre64
x=-5 is a nono in this equation, so never use it, draw a lint to mark its position and stay away from it, that your VA
amistre64
  • amistre64
line not lint, but if it is in macaroni art, use zitis
anonymous
  • anonymous
so to find the equation of the vertical asymptote of the function f(x)=4/x+5 what is the answer?
anonymous
  • anonymous
i am so confused?
amistre64
  • amistre64
lie down on my couch right here and lets talk about this confusion :) tell me, can the bottom of a fraction ever be a zero?
anonymous
  • anonymous
no
amistre64
  • amistre64
then the line on a graph that marks the spot where the bottom would be zero is indictaed by the vertical asymptote. at the value of x that is bad.
amistre64
  • amistre64
what value of x makes the bottom of your fraction there go bad?
anonymous
  • anonymous
anything negative?
amistre64
  • amistre64
negative numbers are not bad numbers they are just misunderstood :) there is on value in particular that is going to be offlimits: x+5 = 0 when x =?
anonymous
  • anonymous
-5
amistre64
  • amistre64
correct! so that is your VA
amistre64
  • amistre64
x = -5 is the VA
anonymous
  • anonymous
ok i get it now,
anonymous
  • anonymous
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anonymous
  • anonymous
there i graphed in it matlab shows you what is happening in case the visual aspect helps
amistre64
  • amistre64
you got that in crayon? :)
anonymous
  • anonymous
then how do i get horizontal asym to 5x^2-4/(x+1)
amistre64
  • amistre64
we divide the top by the bottom like noremal long division to get an exact answer, but its not a horizontal asymptote.... itll be a line that is slanted
anonymous
  • anonymous
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anonymous
  • anonymous
look at the graph that should help you visualize it.
amistre64
  • amistre64
the right side there seems off, but that might be the zoom limit
anonymous
  • anonymous
ill try to fix it
amistre64
  • amistre64
if we simply do a shortcut and divide all the numbers by x, we get the top to be 5x
anonymous
  • anonymous
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anonymous
  • anonymous
there you go now its a bit better
amistre64
  • amistre64
thats better, but the your asymptotes gone parabolic instead of linear :)
amistre64
  • amistre64
5x^2-4/(x+1) the top and bottom differ by 1 degree, so the asymptote is linear
anonymous
  • anonymous
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anonymous
  • anonymous
sorry i had the equation not copy and pasted wrong lol here is the correct result
amistre64
  • amistre64
yay!! i knew you could do it :)
anonymous
  • anonymous
man im just trying to help him out?
amistre64
  • amistre64
youre doing great....
anonymous
  • anonymous
im pretty sure when you learn these things it helps to look at a graph
amistre64
  • amistre64
it does, i am visual alot, the analysis part still gets my brain tied in knots
anonymous
  • anonymous
sorry i thought you were being sarcastic there
amistre64
  • amistre64
;) wel it was in humour, but nothing cruel or hateful. we all have errors to keep us humble i beleive, i know i have my share
anonymous
  • anonymous
well to be honest i did have the right equation i just got excited using matlab and didn bracket off the exponents properly

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