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| Question 6625:  How would the graph of
  look? Answer by khwang(438)
      (Show Source): 
You can put this solution on YOUR website!  Note f(0) =0,the graph passing the origin. The vertical asymptote is x+1 =0. We see that
 when x-->-1+(right), f(x)-->-oo; when x-->-1-(left), f(x)-->oo;
 When x-->oo(or -oo) , y =f(x)--> 1. the horizontal asymptote is y=1.
 Since f(x) = 1- 1/(x+1)
 f'(x) = 1/(x+1)^2, so f'(x) > 0 for all x (not -1)
 and f is an increasing on (1,+oo) or (-oo, 1)
 
 Also ,plot points (1,f(1)), (-2,f(-2)), etc
 Then you will see the graph.
 Another way:by f(x) = 1- 1/(x+1).
 It means f is the composite of the reflection -1/(x+1) and the translation.
 x--> 1/(x+1) (a hyperbola) --> - 1/(x+1) --> 1- 1/(x+1)
 
 Kenny
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