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CalculusHelp
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How do you find horizontal and vertical asymptotes?
I know that for Horizontal I must find the left and right limits of infinity but I'm not sure about vertical asymptotes.
 one year ago
 one year ago
CalculusHelp Group Title
How do you find horizontal and vertical asymptotes? I know that for Horizontal I must find the left and right limits of infinity but I'm not sure about vertical asymptotes.
 one year ago
 one year ago

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dpaInc Group TitleBest ResponseYou've already chosen the best response.3
for rational functions like \(\large f(x)=\frac{g(x)}{h(x)} \), it's what makes g(x) = 0...
 one year ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.3
*** sorry, h(x) = 0
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
alright well if you have a rational function. you get a vertical asymptote if the denominator is zero
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
or like what dpalnc said in a better way
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
so usually in elementary calculus you just have some polynomal, then you just gotta factor it and find the roots of the polynomial
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
or a trig function or something
 one year ago

CalculusHelp Group TitleBest ResponseYou've already chosen the best response.0
Ohh okay thanks for the help, I appreciate it. Also for the polynomial, when I find the roots , how are those included in my graph?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
if you have y = f(x) then the roots tell you were y=0 it is where the curve crosses the xaxis
 one year ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.3
in addition to the rational function i mentioned, g(x) and h(x) must be relatively prime.....
 one year ago

CalculusHelp Group TitleBest ResponseYou've already chosen the best response.0
Oh I get it now, thanks a lot!
 one year ago
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