How do you graph $y = ln((2x-1)/(x+1))$ ?

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Answer 1 · 2017-01-22T08:33:05

. x-intercept ( y = 0 ) : 1. Vertical asymptotes : $x = 1/2 darr and x = -1uarr$ . Horizontal asymptote : $larr y = ln 2 rarr$ :

As $x to +-oo, y = ln ((2-1/x)/(1+1/x)) to ln 2$ .

As $x to (1/2)_+, y to -oo and$ as

$x to -1_-, y to oo$

See the asymptotes-inclusive Socratic graph.

graph{(ln((2x-1)/(x+1))-y)(y-ln 2)(x+1+.01y)(x-1/2+.011y)=0 [-5, 5, -2.5, 2.5]}

How do you graph y = ln((2x-1)/(x+1)) y = ln((2x-1)/(x+1))?