Pre-Calculus: What transformations occur for f(x)=a^x, f(x)=-a^x?

[ronald.vi6]

I am stuck on the per-calculus question below:

What type of transformation would occur with the following exponential functions?

f(x)=a^x

f(x)=-a^x

The possible answers are as follows:

A. Vertical shrinking
B. Vertical stretching
C. Reflection about the x-axis
D. Reflection about the y-axis

If anyone can give an answer to this and help me understand more about the answer I would certainly appreciate it! Thank you!

 

[ksdhart2]

Often times with problems like this, the trick is to gather as much information as you can from the problem and then synthesize it into something meaningful. In this case, you're given two functions. To help alleviate confusion, let's label the second function g(x) = -a^x. Clearly f(x) and g(x) are the same function, except g(x) is negative. Thus, the question is really asking: How does a negative sign out front change a function? To determine that, let's consider some simpler examples?

Let f(x) = x² and g(x) = -x². How do the graphs differ? What if f(x) = x³ and g(x) = -x³? Now how do the graphs differ? And f(x) = x⁴? Based on this pattern, in general, how do the graphs differ for even powers? How do they differ for odd powers? If the behavior is different, why is it different? Now let's consider a slightly trickier example. Let f(x) = sqrt(x) and g(x) = -sqrt(x). How do the graphs differ? Let f(x) = x - 4 and g(x) = -(x - 4) = 4 - x. How do the graphs differ? Does this give you a feel for what's going on?

You may already have divined the answer, but if not, let's move on to the actual problem. Here f and g are functions only of x, meaning a is some arbitrary constant. So, let's fix a = 2. How do the graphs of f(x) and g(x) differ? What about if a = 3? If a = 4?

 

[ronald.vi6]

Pre-Calculus Question

Thanks for the helpful tips and information. When I substitute values for x in the functions, I realize that the graphs for that function are depicted on the positive and negative side of the x-axis on the graphs. To me I would be more inclined to believe that the answer would be a reflecting transformation about the x-axis. Let me know if this is correct.

 

[ksdhart2]

Thanks for the helpful tips and information. When I substitute values for x in the functions, I realize that the graphs for that function are depicted on the positive and negative side of the x-axis on the graphs. To me I would be more inclined to believe that the answer would be a reflecting transformation about the x-axis. Let me know if this is correct.

Yes, that's correct. A negative sign out front results in the same graph except reflect across the x-axis. Very well done.