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Mathematics 11 Online
OpenStudy (anonymous):

x^2-4 FIND THE HOLES ------ x+1

OpenStudy (anonymous):

is it "holes" or "poles"? if it is poles this thing has a pole at -1 because that is where the denominator is 0. i have never heard of it being called a "hole"

OpenStudy (anonymous):

usually called vertical asymptote

OpenStudy (anonymous):

yea vertical asymptote

OpenStudy (anonymous):

or horizontal

OpenStudy (anonymous):

ok vertical asymptote set denominator = 0 and solve

OpenStudy (anonymous):

so long as the numerator is not also zero there, you have a vertical asymptote

OpenStudy (mathteacher1729):

SeeRed -- have you graphed this function?

OpenStudy (anonymous):

in this case \[\frac{(x+2)(x-2)}{x+1}\]

OpenStudy (anonymous):

^yes

OpenStudy (anonymous):

denominator is 0 if x = -1 so it will have a vertical asymptote there. no horizontal asymptote because the degree of the numerator is bigger than the degree of the denominator

OpenStudy (anonymous):

do you need to find the 'slant' asymptote as well?

OpenStudy (anonymous):

no i have the the answer. thak for your help

OpenStudy (anonymous):

welcome

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