How do you graph #y=-2/(x+3)-2# using asymptotes, intercepts, end behavior?
1 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
graph{-1/x [-10, 10, -8, 8]}
2) scaled by a factor of
graph{-2/x [-10, 10, -8, 8]}
2) translated
graph{-2/(x+3) [-10, 10, -8, 8]}
3) translated
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
Hence, we have a vertical asymptote at
Intercepts
When:
# x=0 => y = 2/3-2 = -4/3 #
# y=0 => -1/(x+3)=1 => x=-4#
Hence, the intercepts are
Which is enough the sketch the curve.