SOLUTION: R(t) = 8 - (3)/(2x+4) Please explain how I would go about finding the graph for this. Please give as many details as possible. I know that the vertical asymptote is x=-2, an

Algebra ->  Graphs -> SOLUTION: R(t) = 8 - (3)/(2x+4) Please explain how I would go about finding the graph for this. Please give as many details as possible. I know that the vertical asymptote is x=-2, an      Log On


   

Question 917230: R(t) = 8 - (3)/(2x+4)
Please explain how I would go about finding the graph for this. Please give as many details as possible.
I know that the vertical asymptote is x=-2, and I guess there is no horizontal asymptote because there is no x's in the numerator.
How would I start finding points on the x-axis and overall figuring this out?
Thanks!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
as x gets really really big, and heads off to infinity, 2x+4 blows up to infinity.

So we have 3/(infinity) more or less as x approaches infinity.

The whole fraction - (3)/(2x+4) approaches 0 as x ---> infinity.

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So as x ---> infinity, 8 - (3)/(2x+4) turns into 8 - 0 more or less. That turns into 8.

So the horizontal asymptote is y = 8.

This same logic applies to negative infinity as well.


Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at jim_thompson5910@hotmail.com
or you can visit my website here: http://www.freewebs.com/jimthompson5910/home.html

Thanks,

Jim