ChrisV
graph the fuction f(x)= 1/x+8
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mattt9
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when x goes to zero y approaches 8
ChrisV
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yea i am trying to determine what the graph would look like
mattt9
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it's 1/x not (1/2)x there chief
ChrisV
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well it 1/(x+8)
ChrisV
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but (x-8) is different from x+8 when graphing
mattt9
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you know anyone can type numbers in to wolfram
ChrisV
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(x+8) oops
mattt9
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i'm talking to chief
ChrisV
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yea i want to know how to graph the thing
MasterChief
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Well of course, although I could rule the Earth if it was off limits to everyone but me...
ChrisV
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i can go to a graphing calculator and not come here but it doesnt teach me how to graph it
mattt9
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i disagree chief. conversation over.
ChrisV
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So i assume that I would substitute random numbers for X to graph it?
MasterChief
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Make a table and input numbers for x like 0, and then see what comes out as y. Keep getting numbers until you can make a decent graph.
ChrisV
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so if x= 0 y would = 1/8
MasterChief
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Disagree with what? Obviously, no s**t anyone can use it! I linked a page to the graph, to make things easier even though I misunderstood the question.
MasterChief
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Umm, Well, you put 0 in for x. This cancels out the numerator of 1 above it. The answer would be 8.
MasterChief
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Because 1 divided 0 is nothing, 0.
ChrisV
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1/(x+8)
ChrisV
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i cant figure out how to put the one over both in here
ChrisV
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in the original problem there is no parenthesis
MasterChief
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Oh, that's a different story. Yes, it would com eout as \[1/8\]
ChrisV
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because x+8 and (x+8) would be different shifts when graphing wouldnt it?
MasterChief
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Yes, the solution graph would differ.
ChrisV
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one would shift the graph up and one would shift the graph left
ChrisV
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so i am assuming there would be a hole in my graph at -8
ChrisV
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because -8+8 = 0
MasterChief
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I entered the two equations.
ChrisV
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the first one is correct. Thats where I was confused
ChrisV
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because at x=-9
MasterChief
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Yes, it does move the graph to the right if you seclude the +8
ChrisV
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the equation would result in a negative
ChrisV
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hence a verticle asymtope at -8
ChrisV
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x can never be -8 )
MasterChief
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I would love to answer your question here but I really don't know. I'm in eighth grade and enrolled in Algebra 2. Sorry!
ChrisV
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you did great, and you helped me answer a college calculus question
ChrisV
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just remember that the denominator can never = 0 so if it does you have a hole
ChrisV
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hence for 1/(-8+8)
ChrisV
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=1/0
MasterChief
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Oh wow, Thank you!
And yes, because you can't have, for instance, 4 slices of nothing, right?
ChrisV
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the value doesnt exist
MasterChief
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Like, 4/0
ChrisV
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so in your graph that spot would be left open
ChrisV
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it doesnt exist
MasterChief
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Yes, I know what you mean. It's like trying to have one piece of nothing. That doesn't exist!
MasterChief
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I understand now.
ChrisV
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so in this case we have an asymptope which means we can get closer and closer to 8
MasterChief
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Yup, I get it now thanks!
ChrisV
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such as 7.999
ChrisV
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so the graph goes up and down at 8 infinitely
MasterChief
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I think so.