# Draw a sketch graph of the function y = 3x / (1-x) indicating all the points of interest.

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Sketch `y=(3x)/(1-x)` :

(1) Domain is `RR-{1}` (`x!=1` ) ; there is a vertical asymptote at x=1

(2) The range is `RR-{-3}` (y=-3 ==> `-3=(3x)/(1-x)==>3x-3=3x` ), so there is a horizontal asymptote at y=-3.

(3) `y'=3/((1-x)^2)` (using the quotient rule: `y'=(3(1-x)-3x(-1))/((1-x)^2)`

Since `y'>0` for all x in the domain, the function is increasing on its domain. `y'!=0` for any x, and `y'` fails to exist only at x=1 which is not in the domain, so there are no extrema.

(4) `y''=6/((1-x)^3)` (again using the quotient rule) `y''<0` for `x>1` and `y''>0` for `x<1` , so the curve is concave up on `(-oo,1)` and concave down on `(1,oo)` . There is no inflection point.

(5) `lim_(x->-oo)` is of the form `(-oo)/(oo)` : applying L'Hospital's rule we get `lim_(x->-oo)(3x)/(1-x)=lim_(x->-oo)3/(-1)=-3` as we saw in (2). Simarily, `lim_(x->oo)(3x)/(1-x)=-3` .

(6) The graph is continuous on its domain (it is a rational function).

(7) The y-intercept is at 0 since f(0)=0; the x-intercept is 0 since `0=(3x)/(1-x)==>3x=0` .

The graph:

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