The purpose of this section is to gain experience distinguishing between linear and
exponential behavior,
by recognizing the defining behaviors from small tables of input/output pairs.
Recall from the previous section:




$\Delta y = 4\,$ Linear! $\displaystyle\text{slope} = \frac{\Delta y}{\Delta x} = \frac{4}{1} = 4$ The point $\,(0,3)\,$ tells us that $\,b = 3\,$. Thus, $\,g(x) = 4x + 3\,$. Checking all points: $$ \begin{align} 4(\color{red}{2}) + 3 &= \color{blue}{5}\cr 4(\color{red}{1}) + 3 &= \color{blue}{1}\cr 4(\color{red}{0}) + 3 &= \color{blue}{3}\cr 4(\color{red}{1}) + 3 &= \color{blue}{7}\cr 4(\color{red}{2}) + 3 &= \color{blue}{11} \end{align} $$ 
Be aware that there are infinitely many functions with graphs that pass through
any finite set of data points!
In these exercises, you're finding a plausible function that could generate the data.
The sketch below shows the line found in the previous example, together
with another curve that passes through all the points.
Have fun ‘connecting the dots’ to get many others!
Here's a second example, with the solution written compactly.
In this example, you must solve for the constant $\,b\,$, since the
$\,y\,$intercept is not given in the table.
compact solution  comments  the table of data  
$\Delta x = 3\,$ $\Delta y = 6\,$ $\displaystyle\text{slope} = \frac{\Delta y}{\Delta x} = \frac{6}{3} = 2$ $\,h(x) = 2x + b\,$ Use $\,(2,3)\,$ to solve for $\,b\,$: $3 = 2(2) + b \ \ \Rightarrow\ \ b = 1$ So, $\,h(x) = 2x  1\,$. (Mentally check all points.) 
Be sure to check that $\,\Delta x = 3\,$ for all $\,x\,$values:
$$
2  (1) = 3,\ \ \ \ \ 5  2 = 3,\ \ \ \ \ 8  5 = 3
$$
Be sure to check that $\,\Delta y = 6\,$ for all $\,y\,$values: $$ 3  (3) = 6,\ \ \ \ \ 9  3 = 6,\ \ \ \ \ 159 = 6 $$ Any point can be used to solve for $\,b\,$, so choose the easiest to work with. Recall that ‘$\,\Rightarrow\,$’ is read as ‘implies’. 


$y$values are NOT equally spaced (not linear) Multiplying by $\,9\,$ takes us from $\,y\,$value to successive $\,y\,$value, so it is exponential behavior. Let $\,f(x) = Ka^x\,$. $2 = Ka^0\ \ \ \Rightarrow\ \ K = 2$ $18 = 2a^2\ \ \Rightarrow\ \ a = 3$ Thus, $\,f(x) = 2\cdot 3^x\,$. Checking all points: $$ \begin{align} 2\cdot 3^{\color{red}{2}} &= \color{blue}{\frac 29}\cr 2\cdot 3^{\color{red}{0}} &= \color{blue}{2}\cr 2\cdot 3^{\color{red}{2}} &= \color{blue}{18}\cr 2\cdot 3^{\color{red}{4}} &= \color{blue}{162} \end{align} $$ 
Of course, reallife data usually isn't perfect, like the data sets you're working with in this web exercise. For example, even if, in some experiment, theory predicts that outputs $\,y\,$ should be twice as big as inputs $\,x\,$, your data points probably won't all satisfy the relationship $\,y = 2x\,$ perfectly. Instead of a perfect data point like $\,(1,2)\,$ (where the $\,y\,$value is exactly twice the $\,x\,$value), you might have something like $\,(1\,,\,2.1)\,$ or $\,(0.9\,,\,1.7)\,$. There might be some noise. There might be some error.
In real life, one way to recognize linear or exponential behavior from points in a data set is by plotting them
and observing the graph that results. 

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