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Assume we can write 1/(y+3)=a(x+b) where a and b are constants. Plug in the given points:

(2,-2): 1=a(2+b) and (3,1): ¼=a(3+b). Divide these two equations to remove a:

4=(2+b)/(3+b), 12+4b=2+b, 3b=-10, b=-10/3.

Substitute for b: 1=a(2-10/3)=-4a/3, a=-¾. Therefore 1/(y+3)=-¾(x-10/3).

So inverting both sides: y+3=1/(-¾x+10/4)=4/(10-3x), and y=4/(10-3x)-3, y=(4-30+9x)/(10-3x).

Therefore y=(9x-26)/(10-3x) has an asymptote at y=-3 and passes through (3,1) and (2,-2). It also has an asymptote at x=10/3.

Note the graph has two parts and the asymptotes are common to both parts at y=-3 and x=3⅓. The left component is similar to the graph submitted in the question. The vertical asymptote is shown on the graph above but it is not part of the graph itself.

by Top Rated User (645k points)

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