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

1 Answer
Sep 14, 2017

Answer:

graph{-2/(x+3)-2 [-10, 10, -8, 8]}

Explanation:

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.