How do you graph #y=3^x# and #y=3^(x+1)# and how do the graphs compare?

1 Answer
Dec 25, 2016

See explanation.

Explanation:

The easiest way to graph exponential functions is to simply find a few #(x,y)# pairs that are easy to plot.

Let's start with #y=3^x#.
When #x=0# (i.e. when we're on the #y#-axis), #y=3^0=1#. So the point #(0,1)# is on our graph. From here, every time #x# goes up by #1#, #y# will triple. That is, for every one step right, our step up is 3 times larger than it was before.

#x=1" "=>" "y=3*1=color(blue)3#
#x=2" "=>" "y=3*color(blue)3=color(green)9#
#x=3" "=>" "y=3*color(green)9=27#
and so on.

graph{3^x [-17.15, 8.16, -1.75, 10.9]}

For the graph of #y=3^(x+1)#, we can actually recognize this as a translation of the first graph. The #x#-value is being increased by #1#; what this means is, at every #x#-value, #y=3^(x+1)# looks to the right of #y=3^x# by one unit, finds the #y#-value there, and makes that its own #y#-value at the same #x#. This results in the original graph of #y=3^x# getting shifted to the left by one unit.

Think of it this way: when you look to your left, what was in your field of vision "moves right". Similarly, when an #x#-input decreases by #h# (a.k.a. #(x-h)#), the parent graph shifts right by #h#. That's why all these math formulas like #y=asin[b(x-h)]+k# and #y=a(x-h)^2+k# have "#-h#" instead of "#+h#": because shifting a graph to the right requires a decrease in #x# input, not an increase.

In short: #y=3^(x+1)# will look identical to the graph of #y=3^x#, just shifted one unit to the left.

graph{3^(x+1) [-17.15, 8.16, -1.75, 10.9]}