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how do i find the vertical asymptote?
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do u have a rational function ?
yea its f(x)=x-2/x^2-5x+6
a vertical asymptote or a hole occurs when denominator approaches 0
start by factoring he denominator
x^2-5x+6 could u factor this
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(x-3)(x-2)
so we have a common factor !
\( f(x) = \frac{x-2}{(x-2)(x-3)} \)
\((x-2) \) cancels top and bottom, right ?
yes
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and denominator becomes 0, when x = 2, or x = 3
but since (x-2) is getting cancelled, it is just a hole. its not a vertical asymptote since (x-3) is not getting cancelled, it is a proper vertical asymptote
x-3 = 0 x = 3 we say the line x = 3 is the vertical asymptote of rational function
thanks i understand now
glad to hear :)
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