How to graph by hand: y=log((x/(x+2))

  • Thread starter srfriggen
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  • #1
srfriggen
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Homework Statement:
Sketch a graph, without the use of a calculator, y=log((x/(x+2))
Relevant Equations:
y=log((x/(x+2))
I first attempted to find the x and y intercepts, algebraically, and discovered there were none. I then split the equation into y= log(x) - log(x+2) to see if that would give me any insight. It did not.

I used a graphing calculator and saw many similarities between x/(x+2) and log((x/(x+2)) but cannot make the leap to the latter.

I'm now considering the end behavior for the inner function and realizing there is a horizontal asymptote of y = 1. That means the end behavior of the outer function must have a horizontal asymptote at 0, since we are evaluating numbers closer and closer to 1 with the log function.

There is a vertical asymptote as x = -2

I think now I just need to test some points to see where the function is positive or negative.

And I just realized there must be an asymptote where the inner function is zero, i.e. x = 0. So I just tested some points and found the answer.
 
Last edited:

Answers and Replies

  • #2
pasmith
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It may be easier to plot [itex]x = f(y)[/itex] first. But that's only because I know [itex]f(y)[/itex] is a translation and scaling of a hyperbolic function which I can plot without a calculator.

Otherwise, remember that [itex]\log[/itex] is only defined for positive arguments, and is positive when its argument is greater than 1.
 
  • #3
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As an alternative to what @pasmith wrote, I would sketch (by hand) a graph of y = x/(x + 2), noting that there is an obvious vertical asymptote around x = -2. Also, because this function is the quotient of two polynomials of the same degree, there will be a horizontal asymptote. Once you have figured out what the graph looks like around the vertical asymptote and for large or very negative x values, you should have a reasonable sketch of this function.

Then, do a graph of y = log(x/(x + 2)), keeping in mind that this function is defined only where x/(x + 2) > 0.
 
  • #4
anuttarasammyak
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[tex]y=-\log(1+\frac{2}{x})[/tex]
How about drawing graph of
[tex]y=1+\frac{2}{x}[/tex]at first to know for which x y>0, y=0, y=1, y=##\infty##.
 
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