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

[srfriggen]

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.

 

[pasmith]

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

Otherwise, remember that \log is only defined for positive arguments, and is positive when its argument is greater than 1.

 

[Mark44]

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.

 

[anuttarasammyak]

y=-\log(1+\frac{2}{x})
How about drawing graph of
y=1+\frac{2}{x}at first to know for which x y>0, y=0, y=1, y=$\infty$．