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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you first cross out like terms from top to bottom, then what ever value makes the bottom a zero is your VA

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0in this case....x =5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you just make x+5=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try finding the lim as x goes to 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that is correct, you want to know what vale of x can never be used

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x=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
 5 years ago
Best ResponseYou've already chosen the best response.0line not lint, but if it is in macaroni art, use zitis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so to find the equation of the vertical asymptote of the function f(x)=4/x+5 what is the answer?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lie down on my couch right here and lets talk about this confusion :) tell me, can the bottom of a fraction ever be a zero?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then 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
 5 years ago
Best ResponseYou've already chosen the best response.0what value of x makes the bottom of your fraction there go bad?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0negative 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 =?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0correct! so that is your VA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there i graphed in it matlab shows you what is happening in case the visual aspect helps

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you got that in crayon? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then how do i get horizontal asym to 5x^24/(x+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0look at the graph that should help you visualize it.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the right side there seems off, but that might be the zoom limit

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if we simply do a shortcut and divide all the numbers by x, we get the top to be 5x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there you go now its a bit better

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats better, but the your asymptotes gone parabolic instead of linear :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05x^24/(x+1) the top and bottom differ by 1 degree, so the asymptote is linear

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i had the equation not copy and pasted wrong lol here is the correct result

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yay!! i knew you could do it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0man im just trying to help him out?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0youre doing great....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im pretty sure when you learn these things it helps to look at a graph

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it does, i am visual alot, the analysis part still gets my brain tied in knots

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i thought you were being sarcastic there

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0;) 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
 5 years ago
Best ResponseYou've already chosen the best response.0well 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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