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

1 Answer
George C.
Aug 30, 2015

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.

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