Introduction

Welcome back to AP Precalculus! In the previous article, we discussed how to find the change in arithmetic and geometric sequences. In this article, we’ll be discussing something very similar, but instead for linear and exponential functions. Without further adieu, let’s get started!

Linear functions

Consider the following function.

$y=3x+5$

You’ve probably seen functions just like the one above in math class since algebra. As you should also remember from algebra, this function has a slope of $3$ and a $y$-intercept of $5$. This is called a linear function. A linear function is a function that has a constant rate of change. For the function above, the function always has a rate of change of $3$ over its domain.

As you learned in topic 2.1, arithmetic sequences can have a similar form. In this case, the equivalent form would be 

$A_n=a_0+3n$

$a_0=5$

In this case, the arithmetic sequence above is identical to our original function, when only considering non-negative integers. In contrast, our linear function above is defined for all real numbers. They are the same, but they have different domains.

Similar to arithmetic sequences, we are also able to create a linear function even if we don’t know the $y$-intercept, or in the case of an arithmetic sequence, the $0$th term. If all we know is the slope of the line and a specific point on the function, we are able to create a new function based only off of that information. This is called the point-slope form of the function.

$y=m(x-x_i)+y_i$

where $(x_i,y_i)$ is a point on the line of the function and $m$ is the slope of the function.

For example, let’s consider a linear function that has a slope of $2$ and passes through the point $(5,3)$. We can plug our values into the point-slope form and solve to get our equation in standard form.

$y=m(x-x_i)+y_i$

$y=2(x-5)+3$

$y=2x-10+3$

$y=2x-7$

Plugging in $x=5$ to check, we get $y=2(5)-7=10-7=3$, which aligns with the above information. 

Exponential functions

Exponential functions can be treated in a similar way to geometric sequences similarly to the relationship between arithmetic sequences and linear functions. Consider an exponential function g(x).

$g(x)=3\cdot 5^x$

The equivalent form as a geometric sequence would be as follows.

$g_n=g_n-1\cdot 5$

$g_0=3$

Likewise, we are able to create an exponential function if we know just the ratio and a point.

$g(x)=y_i\cdot r^{x-x_i}$

where $(x_i,y_i)$ is a point on the line of the function and $r$ is the ratio.

As an example, let $g(x)$ be an exponential function with a ratio of $3$ that passes through the point $(4,15)$. We can use our modified point-slope equation to get the equation of $g(x)$ in standard form.

$g(x)=15\cdot 3^{x-4}$

$g(x)=15\cdot 3^x\cdot 3^{-4}$

$g(x)=15\cdot 3^x\cdot \frac{1}{81}$

$g(x)=\frac{15}{81}\cdot 3^x$

$g(x)=\frac{5}{27}\cdot 3^x$

To check, let’s calculate $g(3)$.

$g(3)=\frac{5}{27}\cdot 3^3$

$g(3)=\frac{5}{27}\ast \cdot 27$

$g(3)=5$

Domains

As mentioned prior, a sequence and a function do not have the same domain. Take the following arithmetic sequence and linear function.

$f(x)=3x+5$

$a_n=3+a_n-1$

$a_0=5$

We’re able to calculate $f(3.5)$ or $f(-2)$, but we can’t get the $3.5$th or $-2$nd element of a sequence; that simply doesn’t make sense. The domain of a sequence and its corresponding function will differ. The domain of an exponential function will be all real numbers, but the domain of a geometric sequence will be only non-negative integers.

Distinguish between linear and exponential functions based on inputs and outputs

(Note, remember this phrasing, this will be part of your FRQ)

Given a table of inputs and outputs, you can tell if a function is linear if, over equal-length input values, the output values change at a constant rate. You can tell if a function is exponential if, over equal-length input values, the output values change proportionally.

What does this mean? Consider the two tables below.

$\begin{array}{ccccc}x& 1& 2& 4& 6\\ f(x)& 4& 7& 13& 19\end{array}$  

$\begin{array}{ccccc}x& 1& 2& 4& 6\\ f(x)& 4& 8& 32& 128\end{array}$  

Let’s look at the first table. You can see that, if we are to take any two input-output pairs, we can use the linear rate of change formula, that being

$\frac{[f(x_2)-f(x_1)]}{(x_2-x_1)}$

to get the rate of change of the function $f(x)$. No matter what two input-output pairs you use the formula on, you will always get the same result.  If we were to calculate the average rate of change along the interval $[1,2]$, we’d get $\frac{f(2)-f(1)}{2-1}=\frac{7-4}{2-1}=\frac{3}{1}=3$, which would be the same as if we calculated the rate of change along the interval $[2,6]$: $\frac{f(6)-f(2)}{6-2}=\frac{19-7}{6-2}=\frac{12}{4}=3$

Now let’s take a look at the second table. We can’t use our linear rate of change formula on this, as the average rate of change between input-output pairs differs. If we were to try to calculate the average rate of change between two intervals, we will get differing results.

$\frac{g(2)-g(1)}{2-1}=\frac{8-4}{2-1}=\frac{4}{2}=2$

$\frac{g(6)-g(2)}{6-2}=\frac{128-8}{6-2}=\frac{120}{4}=30$

We can instead use the exponential rate of change formula, that being 

$\sqrt[\Delta x]{\frac{g(x_2)}{g(x_1)}}$ with $\Delta x=x_2-x_1$

Now, no matter which two input-output pairs we choose, we will still get the same result.

$\Delta x=2-1=1$

$\sqrt[1]{\frac{g(2)}{g(1)}}=\sqrt[1]{\frac{8}{4}}=\frac{8}{4}=2$

and

$\Delta x=6-2=4$

$\sqrt[4]{\frac{g(6)}{g(2)}}=\sqrt[4]{\frac{128}{8}}=\sqrt[4]{16}=2$

Practice Questions