Given #h(x)=e^-x#, how do you describe the transformation?

1 Answer
Jun 6, 2017

It is a reflection across the y-axis of #f(x)=e^x#

Explanation:

I'm going to assume you mean the transformation from the parent function #f(x) = e^x#.

Let's think about how to get #e^-x# from #f(x)#:

You would have to plug in #-x# instead of #x#, giving #f(-x)=e^-x#

Therefore we can say that:

#h(x) = f(-x)#

What does this mean as far as transformation on a graph? Well...

The point #(2, e^2)# on #f(x)# will correspond to the point #(-2, e^2)# on #h(x)#.

In fact, plugging in any value into f(x) and then plugging in the negative of that value to h(x) will give the same answer.

This basically means that the x-values of #f(x)# become the negative of what they were originally.

Graphically, this means the graph of #f(x)# is reflected across the y-axis to get #h(x)#.

You can see this below: #e^x# and #e^-x# are graphed together to show how #e^-x# is a reflection of #e^x#.

#color(white)"XXXXXXXXXXXXXX-"e^-x color(white)"XXXXX"e^x#
#color(white)"XXXXXXXXXXXXXX-"darr color(white)"XXXXX"darr#
graph{(y-e^x)(y-e^-x)=0 [-10.16, 9.84, -3.4, 6.6]}