How do you graph #y = 4^x-2#?

1 Answer
May 18, 2016

Please see below.

Explanation:

To graph any function, we should have a broad idea about the function. As far as possible choose points carefully. Initially it makes good sense to pick up points where #x# or #y# are zero. These give points on the two axis.

In function #y=4^x-2#, #x=0# gives #y=-2# and #y=0# gives #4^x=2# or #x=log_(4)2=log2/log4=log2/2log2=1/2#. Hence, two pints through which the curve passes are #(0,-2)# and #(1/2,0)#.

Further as #x# appears as a power of #4# in #y=4^x-2# as #x->-oo# #y=-2# hence we have a horizontal asymptote as #y=-2#.

We can also find a few other values, by putting #x# as #{-1,1,3/2,2,5/2,3}# etc. We have use #2# in denominator as it is equivalent to square root of #4#, easy to calculate. Values of #x# are chosen narrowly spaced small values as too large values will lead to too large values of #y#.

Using these values of #x#, we get #y# as #{-7/4,2,6,14,30,62}# and points are

#(-1,-7/4)#, #(1,2)#, #(3/2,6)#, #(2,14)#, #(5/2,30)# and #(3,62)#. WE have already got #(0,-2)# and #(1/2,0)#. Further on left hand side the curve should tend to the value #-2# as we have an asymptote.

Hence, the curve should be as shown below..

graph{y=4^x-2 [-10, 10, -5, 5]}