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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