Linear functions ($\,y = mx + b\,$) and
exponential functions ($\,y = b^x\,$) are both
extremely important.
However, they behave very
differently, and it's important that you thoroughly understand this difference.
In particular, you should be able to recognize linear and exponential behavior when looking at tables of input/output pairs which have equallyspaced inputs; this is covered in the next section.
Here's what you need to think about for this lesson:
moving from left to right. You stop at a point. Then, you move a fixed amount in the horizontal direction. How does your ‘elevation’ change? That is, how do you change vertically? Keep walking. Stop at a different point. Move the same horizontal amount. Now, how much do you change vertically? Repeat. 

For general functions, you can't say much about your vertical movements. For linear and exponential functions, however, your up/down movement is completely predictable! 
Firstly, let's firm up the notation. For the graph of a given function $\,f\,$:

In the earlier section Working with Linear Functions, we learned that for a linear function $\,f(x) = mx + b\,$: $$\overbrace{f(x+\Delta x)}^{\text{new $\,y\,$value}} = \overbrace{f(x) + m\Delta x}^{\text{formula involving original $\,y\,$value}}$$
What is that formula on the right telling us about the new $\,y\,$value?
You already know this result well from years of working with linear functions.
However, it is important to review before contrasting with the
very different behavior of exponential functions.
Let's now consider exponential functions. Let $\,f(x) = a^x\,$ for $\,a > 0\,$, $\,a\ne 1\,$. (The graph at right shows an exponential function for $\,a > 1\,$.) The function $\,f\,$ takes an input and raises $\,a\,$ to that power. Then, $$ \begin{alignat}{2} f(x+\Delta x) &= a^{x + \Delta x} \qquad&&\text{(definition of $\,f\,$)}\cr\cr &= a^xa^{\Delta x}&&\text{(exponent law)}\cr\cr &= a^{\Delta x}\cdot f(x)\qquad&&\text{(since $\,f(x) = a^x\,$)} \end{alignat} $$ Thus, for exponential functions: $$ f(x+\Delta x) = a^{\Delta x}\cdot f(x) $$ What is that formula on the right telling us about the new $\,y\,$value?

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
