How to sketch #y=2^(x+2) -1?# - Detailed explanation please

1 Answer
Jan 20, 2018

Please see below.

Explanation:

.

It is easier if we graph #y=2^x# as the base function first. This is an exponential function. To do this, we pick a few values for #x# and plug them into the function to find the corresponding #y# for each one:

#x=1#, #y=2^1=2#

#x=2#, #y=2^2=4#

#x=3#, #y=2^3=8#

#x=oo#, #y=2^oo=oo#

#x=-1#, #y=2^-1=1/2^1=1/2#

#x=-2#, #y=2^-2=1/2^2=1/4#

#x=-3#, #y=2^-3=1/2^3=1/8#

#x=-oo#, #y=2^-oo=1/2^oo=1/oo=0#

These results indicate that as #x# goes to #oo#, #y# goes to #oo#, and as #x# goes to #-oo#, #y# goes to #0#, which means the #y#-axis is the horizontal asymptote. Plotting the points on a coordinate system gives us the graph:

graph{2^x [-10, 10, -5, 5]}

We know that if we add a constant to our #x# the graph shifts in the negative direction by that number of units.

We also know that if we subtract a constant from our #y# the graph will shift down by that number of units.

Therefore, if we add #2# to our #x# and subtract #1# from our #y# the graph will shift #2# units to the left and #1# unit down, and our function becomes:

#y=2^(x+2)-1#

The graph of this function is:

graph{2^(x+2)-1 [-10, 10, -5, 5]}