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Mathematics 18 Online
OpenStudy (anonymous):

identify the asymptote: (1/x^2) + 3

OpenStudy (anonymous):

The Possible answers are: x=2,y=-3 x=2,y=3 x=-2,y=-3 x=-2,y=3

OpenStudy (anonymous):

i disagree with all the possible answers

OpenStudy (anonymous):

well those are the ones there are

OpenStudy (anonymous):

well they're wrong

OpenStudy (anonymous):

x=2 and x=-2 are not asymptotes simply because if you substituted them in, you would get an answer

OpenStudy (anonymous):

ignoring the 3 for now cause all that will do is shift hte graph up 3 units and looking at 1/x^2 recall that the denominator cannot be equal to 0 that means x cannot be _____ and that value will be the x- asymptote

OpenStudy (anonymous):

so its either c or d

OpenStudy (anonymous):

no, as i said, none of the answer choices are correct

OpenStudy (anonymous):

well oneof them must be close to the correct answer

OpenStudy (anonymous):

none of them are correct or you did not provide the full question

OpenStudy (anonymous):

only 2 are partly correct and neither of them are any close to the answer

OpenStudy (anonymous):

here is the full question Identify the asymptotes of the graph: y=(1/x+2) + 3

OpenStudy (anonymous):

thts what it says

OpenStudy (anonymous):

that is a significantly different question than what is posted above

OpenStudy (anonymous):

oh ya duh my mistake srry

OpenStudy (anonymous):

for the parent graph 1/x the x and y asymptote would both be zero 1/x +b y asymptote would be zero and y asymptote would be "b" because the values/graph will not be able to cross y=b to determine x asymptote the denominator cannot be equal to zero so you set the denominator equal to zero and solve for x and x will be the x asymptote

OpenStudy (anonymous):

So its B then

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