Exponential graphs: If y=2^-x is graph of y=2^x reflected in the y axis why isn't...

[MEC]

If y=2^-x is graph of y=2^x reflected in the y axis why isn't y=2^[-(x-1)] just the graph of y=2^(x-1) reflected in the y axis? Thanks.

 

[Deleted member 4993]

If y=2^-x is graph of y=2^x reflected in the y axis why isn't y=2^[-(x-1)] just the graph of y=2^(x-1) reflected in the y axis? Thanks.

When you are reflecting about y-axis, you are actually reflecting about the line x=0

For transformation of 2^(x-1) to 2[-(x-1)] you will need to reflect about x-1 = 0 line.

 

[MEC]

Thanks for the reply. I see from graphing that is the line I'd need to reflect in but I don't understand why. Could you explain please. Thanks.

 

[Deleted member 4993]

If y=2^-x is graph of y=2^x reflected in the y axis why isn't y=2^[-(x-1)] just the graph of y=2^(x-1) reflected in the y axis? Thanks.

Thanks for the reply. I see from graphing that is the line I'd need to reflect in but I don't understand why. Could you explain please. Thanks.

Do you understand - why the process of transformation from \(\displaystyle 2^{(x-0)} \to 2^{-(x-0)}\) involves reflection about the line x=0?

 

[MEC]

No. If I'm being honest I've just rote learned that y=2^-x is just y=2^x reflected over the y axis and assumed until I graphed it that the function I posted about would work in the same way.

 

[MEC]

No sorry. I think I've just rote learned that y=2^-x is y=2^x reflected in the y axis and assumed that the function I originally posted about would reflect in the same way.

 

[Steven G]

Thanks for the reply. I see from graphing that is the line I'd need to reflect in but I don't understand why. Could you explain please. Thanks.

Let x be 1 more and 1 less than 1 (so x will be 2 or 0). What will x-1 be for each of these two x's?
Now let x be 2 more and 2 less than 1. What will x-1 be for each x?
Now let x be 3 more and 3 less than than 1. What will x be?
Do YOU see the symmetry??
Please respond to my questions. Good luck.