How to find the asymptotes of #f(x)= 0.72^x -2#?

1 Answer
Mar 26, 2016

#y=-2#
I have invited another mathematician to check this

Explanation:

Given:#" "f(x)=0.72^x-2#

There are three prime conditions for this expression

Condition 1: #x>0#

Condition 2: #x<0#

Condition 3: #x=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider condition 3 ")x=0#

When #x=0" "f(x)# is defined as it has a definite value.

Thus #color(blue)(x=0 " is NOT an asymptote")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider condition 2 ") x<0#

The #f(x)# takes the form #1/(0.72^x)-2#

When #x=0": "f(x)=1-2" "# thus defined

As #x# becomes bigger #0.72^2# becomes smaller so

#1/(0.72^2)" "#becomes bigger

Thus #lim_(xtooo) f(x)->oo# (infinity is not a number!)

#color(brown)("Is this an asymptote? I have invited another mathematician to look at this bit!")#
Gut feeling is that this is NOT an asymptote!!!!!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider condition 1 ") x>0#

As #x# becomes increasingly larger #0.72^x# becomes increasingly smaller and smaller.

So# lim_(xtooo) 0.72^x-2 -> -2color#

#color(blue)(-2" Is an asymptote")#