How do you graph #y=1.1(0.1)^x#?

1 Answer
Aug 30, 2015

Answer:

Note that the graph is exponentially decaying and find some points through which it goes, to find that it is very steep for negative values of #x# and very flat for positive values.

Explanation:

This graph is very steep for negative values of #x# and very flat for positive values of #x#, passing through the points:

#(-2, 110)#, #(-1, 11)#, #(0, 1.1)#, #(1, 0.11)#, #(2, 0.01)#

graph{1.1*(0.1)^x [-5.23, 4.77, -0.95, 4.05]}

Note that reversing the sign of #x# or replacing the #0.1# with #10# results in the mirror image graph:

graph{1.1*(10)^x [-5.23, 4.77, -0.95, 4.05]}

For practical purposes, it is often more useful to graph the common logarithm of the function instead,

#log(y) = log(1.1(0.1)^x)) = log(1.1)+x log(0.1)#

#= log(1.1)-x ~~ 0.04139-x#

graph{log(1.1(0.1)^x) [-2, 3, -1.12, 1.38]}

Graphing the #log# of the function is basically the same as graphing the original function on paper with a logarithmic vertical scale.