- anonymous

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

- katieb

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- anonymous

when x goes to zero y approaches 8

- anonymous

yea i am trying to determine what the graph would look like

- MasterChief

But I am thy Master Chief?
http://www.wolframalpha.com/input/?i=y%3D.5x%2B8

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## More answers

- anonymous

it's 1/x not (1/2)x there chief

- anonymous

well it 1/(x+8)

- MasterChief

Preposterous! How could I have made such a delinquent mistake?
http://www.wolframalpha.com/input/?i=y%3D%281%2Fx%29%2B8

- anonymous

but (x-8) is different from x+8 when graphing

- anonymous

you know anyone can type numbers in to wolfram

- anonymous

(x+8) oops

- anonymous

i'm talking to chief

- anonymous

yea i want to know how to graph the thing

- MasterChief

Well of course, although I could rule the Earth if it was off limits to everyone but me...

- anonymous

i can go to a graphing calculator and not come here but it doesnt teach me how to graph it

- anonymous

i disagree chief. conversation over.

- anonymous

So i assume that I would substitute random numbers for X to graph it?

- MasterChief

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.

- anonymous

so if x= 0 y would = 1/8

- MasterChief

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

Umm, Well, you put 0 in for x. This cancels out the numerator of 1 above it. The answer would be 8.

- MasterChief

Because 1 divided 0 is nothing, 0.

- anonymous

1/(x+8)

- anonymous

i cant figure out how to put the one over both in here

- anonymous

in the original problem there is no parenthesis

- MasterChief

Oh, that's a different story. Yes, it would com eout as \[1/8\]

- anonymous

because x+8 and (x+8) would be different shifts when graphing wouldnt it?

- MasterChief

Yes, the solution graph would differ.

- anonymous

one would shift the graph up and one would shift the graph left

- anonymous

so i am assuming there would be a hole in my graph at -8

- anonymous

because -8+8 = 0

- MasterChief

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

- MasterChief

I entered the two equations.

- anonymous

the first one is correct. Thats where I was confused

- anonymous

because at x=-9

- MasterChief

Yes, it does move the graph to the right if you seclude the +8

- anonymous

the equation would result in a negative

- anonymous

hence a verticle asymtope at -8

- anonymous

x can never be -8 )

- MasterChief

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!

- anonymous

you did great, and you helped me answer a college calculus question

- anonymous

just remember that the denominator can never = 0 so if it does you have a hole

- anonymous

hence for 1/(-8+8)

- anonymous

=1/0

- MasterChief

Oh wow, Thank you!
And yes, because you can't have, for instance, 4 slices of nothing, right?

- anonymous

the value doesnt exist

- MasterChief

Like, 4/0

- anonymous

so in your graph that spot would be left open

- anonymous

it doesnt exist

- MasterChief

Yes, I know what you mean. It's like trying to have one piece of nothing. That doesn't exist!

- MasterChief

I understand now.

- anonymous

so in this case we have an asymptope which means we can get closer and closer to 8

- MasterChief

Yup, I get it now thanks!

- anonymous

such as 7.999

- anonymous

so the graph goes up and down at 8 infinitely

- MasterChief

I think so.

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