Question 987496
I am going to write this as 6- [1/(x+3)]
That is very different from 6-(1/x) +3
The domain is all reals except x= -3
Range: look at x approaching -3 from the + side and from the - side.
As x approaches from the + side, the fraction gets large positive 1/(-2.999+3)=1/.001=1000, and that makes the function large negative, since it is being subtracted.
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And by the same approach from the negative side, it can be infinite negative as well, which subtracted, makes the range infinitely positive.
It is DISCONTINUOUS at x=-3
It increases from minus infinity to 6 and decreases from positive infinity to 6 as x gets large negative.  The bounds are asymptote is 6. As x gets large positive, the function goes from negative infinity to 6.
It is 1 to 1, for while it doesn't look like it passes either horizontal or vertical line test, the function does change slightly.
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Solving for roots, -1/(x+3)=-6
6x+18=1
6x=-17
x= -(17/6)




{{{graph(300,200,-20,20,-50,50,6-(1/(x+3)))}}}