how do i find the vertical asymptote of the function f(x)=4/(x+5)

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how do i find the vertical asymptote of the function f(x)=4/(x+5)

Mathematics
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you first cross out like terms from top to bottom, then what ever value makes the bottom a zero is your VA
in this case....x =-5
so you just make x+5=0

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then solve for x?
Try finding the lim as x goes to 0
that is correct, you want to know what vale of x can never be used
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
line not lint, but if it is in macaroni art, use zitis
so to find the equation of the vertical asymptote of the function f(x)=4/x+5 what is the answer?
i am so confused?
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?
no
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.
what value of x makes the bottom of your fraction there go bad?
anything negative?
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 =?
-5
correct! so that is your VA
x = -5 is the VA
ok i get it now,
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there i graphed in it matlab shows you what is happening in case the visual aspect helps
you got that in crayon? :)
then how do i get horizontal asym to 5x^2-4/(x+1)
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
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look at the graph that should help you visualize it.
the right side there seems off, but that might be the zoom limit
ill try to fix it
if we simply do a shortcut and divide all the numbers by x, we get the top to be 5x
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there you go now its a bit better
thats better, but the your asymptotes gone parabolic instead of linear :)
5x^2-4/(x+1) the top and bottom differ by 1 degree, so the asymptote is linear
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sorry i had the equation not copy and pasted wrong lol here is the correct result
yay!! i knew you could do it :)
man im just trying to help him out?
youre doing great....
im pretty sure when you learn these things it helps to look at a graph
it does, i am visual alot, the analysis part still gets my brain tied in knots
sorry i thought you were being sarcastic there
;) 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
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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