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graph the fuction f(x)= 1/x+8

Mathematics
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when x goes to zero y approaches 8
yea i am trying to determine what the graph would look like
But I am thy Master Chief? http://www.wolframalpha.com/input/?i=y%3D.5x%2B8

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Other answers:

it's 1/x not (1/2)x there chief
well it 1/(x+8)
Preposterous! How could I have made such a delinquent mistake? http://www.wolframalpha.com/input/?i=y%3D%281%2Fx%29%2B8
but (x-8) is different from x+8 when graphing
you know anyone can type numbers in to wolfram
(x+8) oops
i'm talking to chief
yea i want to know how to graph the thing
Well of course, although I could rule the Earth if it was off limits to everyone but me...
i can go to a graphing calculator and not come here but it doesnt teach me how to graph it
i disagree chief. conversation over.
So i assume that I would substitute random numbers for X to graph it?
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.
so if x= 0 y would = 1/8
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.
Umm, Well, you put 0 in for x. This cancels out the numerator of 1 above it. The answer would be 8.
Because 1 divided 0 is nothing, 0.
1/(x+8)
i cant figure out how to put the one over both in here
in the original problem there is no parenthesis
Oh, that's a different story. Yes, it would com eout as \[1/8\]
because x+8 and (x+8) would be different shifts when graphing wouldnt it?
Yes, the solution graph would differ.
one would shift the graph up and one would shift the graph left
so i am assuming there would be a hole in my graph at -8
because -8+8 = 0
Sorry, it appears so. I have not taken this but look: http://www.wolframalpha.com/input/?i=y%3D1%2F%28x%2B8%29%2C+y%3D%281%2Fx%29%2B8
I entered the two equations.
the first one is correct. Thats where I was confused
because at x=-9
Yes, it does move the graph to the right if you seclude the +8
the equation would result in a negative
hence a verticle asymtope at -8
x can never be -8 )
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!
you did great, and you helped me answer a college calculus question
just remember that the denominator can never = 0 so if it does you have a hole
hence for 1/(-8+8)
=1/0
Oh wow, Thank you! And yes, because you can't have, for instance, 4 slices of nothing, right?
the value doesnt exist
Like, 4/0
so in your graph that spot would be left open
it doesnt exist
Yes, I know what you mean. It's like trying to have one piece of nothing. That doesn't exist!
I understand now.
so in this case we have an asymptope which means we can get closer and closer to 8
Yup, I get it now thanks!
such as 7.999
so the graph goes up and down at 8 infinitely
I think so.

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