How do you graph `y=-2/(x+3)-2` using asymptotes, intercepts, end behavior?

I would advocate that just because we have more advanced mathematical skills , it does not mean that we should automatically use them in favour of basic translation skills.

We can write the given equation:

`y = -2/(x+3)-2`

In the form:

`(y+2) = -2/((x+3))`

We should now recognise this as the graph of the function

`y=1/x`

graph{1/x [-10, 10, -8, 8]}

Which is:

1) inverted to give `y=-1/x`
graph{-1/x [-10, 10, -8, 8]}

2) scaled by a factor of `2` to give `y=-2/x`
graph{-2/x [-10, 10, -8, 8]}

2) translated `3` units to the left to give `y=-2/(x+3)`
graph{-2/(x+3) [-10, 10, -8, 8]}

3) translated `2` units down to give `(y+2)=-2/(x+3)`
graph{-2/(x+3)-2 [-10, 10, -8, 8]}

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In the spirit of the question if we must use asymptotes, intercepts, end behavior then:

`y = -2/(x+3)-2`

End Behaviour:

`x rarr +oo => x+3 rarr +oo => (-2)/(x+3) rarr 0^-`
`x rarr -oo => x+3 rarr -oo => (-2)/(x+3) rarr 0^+`

Hence the end behaviour is as follows:

`lim_(x rarr +oo) y = -2^-` and `lim_(x rarr -oo) y = -2^+`

Asymptotes

The denominator is zero when `x+3 = 0`

Hence, we have a vertical asymptote at `x=-3`

Intercepts

When:

`x=0 => y = 2/3-2 = -4/3`
`y=0 => -1/(x+3)=1 => x=-4`

Hence, the intercepts are `(0,-4/3)` and `(-4,0)`

Which is enough the sketch the curve.