How do you graph #y=2(1/3)^(x+1)-3#?

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Answer 1 · 2017-02-06T08:50:02

Plug and plot

I would just try values.

if #x = -2# then #y= 3#

if #x =-1# then #y= -1#

if #x = 0# then #y= -7/3#

if #x = 1# then #y= -25/9#

Then graph, connecting the points
graph{2(1/3)^(x+1)-3 [-10, 10, -5, 5]}