Page 2 - Mathematics: Algebraic Reasoning Study Guide for the TSIA2

Relations and Functions

Relations and functions are mathematical statements showing how two variables are related to each other. Note that, rather than writing \(y=3x-5\), functions are often written using a symbol \(f(x)\), read “f of \(x\)”. For example, \(f(x) = 3x-5\). \(g(x)\) or \(h(x)\) are also commonly used. You may very well see other variables than \(x\).


A relation between two variables, x and y, states a relationship between the variables, such as \(x \le y\). A relation is true for some x,y pairs such as (3,5) in this example but false for other pairs such as (5,3). In this example the x value must be less than or equal to the y value for the relation to be true.


A function is a specific form of relation where every x value has only one y value that makes the relation true. The previous example, \(x \le y\), is not a function because for any x value, there are many y values that are equal to or greater than x. \(x=y\), however, is a function because only one value of y can be equal to a given value of x. A function is given in the form \(y=f(x)\), where \(f(x)\) is an expression with the variable x. It may also be shown as \(f(x)=(\text{an expression with } x)\).

Linear Functions

Linear functions are functions that have no higher degree than one. For example, \(f(x) = 7x+19\).

Evaluating a Linear Function

To evaluate any function is to substitute a given value for the variable and simplify the result. The problem will look like this: If \(f(x) = 5x -9\), evaluate \(f(2)\).


\[\text{If } f(x) = 4x-27 \text{ evaluate } f(8)\] \[f(8) =4 \cdot 8 - 27\] \[f(8) = 32-27\] \[f(8) = 5\]

Problem-Solving in Context

Many real-life situations can be described with linear models. Often, the information given in a problem can be used to form an equation that describes the situation. This is the model. Usually, the equation can be solved to find the required solution.

For example, suppose a heavy rain is falling at a rate of \(1.6 \;\frac{inch}{hour}\). How long will it take to accumulate \(10.4\) inches? If we let the rate be \(r\), the depth be \(d\), and the time be \(t\), we can write this equation:


Substituting the known quantities, we get:

\[10.4=1.6 \cdot t\]

This is a linear model for the rainfall. We can solve for \(t\).

\[t=\dfrac{10.4}{1.6}\] \[t =6.5 \text{ hours}\]

Exponential Decay and Growth

If a certain quantity increases at a constant rate each time period, that is called exponential growth. For example, a savings account may grow by \(1.25\%\) each year. On the other hand, if a certain quantity decreases by a constant rate each time period, that is called exponential decay. An example of that is the way the value of a computer may decrease at a rate of \(20\%\) per year.

General Procedure

Formulas for exponential growth or decay can be confusing because, depending on where you run into the formulas, you will see different variables and different arrangements used. For this guide we will use these two versions.

\(a=a_0(1+r)^t\) for exponential growth

\(a=a_0(1-r)^t\) for exponential decay

Symbols Key:

\(a\) = amount at the end of the period

\(a_0\) = beginning amount

\(r\) = rate of change as a decimal. \((\)e.g., \(3\% = 0.03)\)

\(t\) = number of time periods elapsed

Example 1

Suppose a bacteriologist introduces 200 bacteria into a growth medium in a petri dish. The number of bacteria is known to increase at a rate of \(30\%\) each hour. How many bacteria will be in the dish after 6 hours?

Because the rate is constant each hour, this is an example of exponential growth, and we will use the formula \(a=a_0(1+r)^t\).

\[a_0 = 200\] \[r = 0.30\] \[t=6\] \[a=200(1+0.30)^6\] \[a = 200(1.30)^6\] \[a = 200(4.827)\] \[a = 965\]

Example 2

The radioactive gas radon-220 decays at a rate of \(1.33 \%\) per second. If you start with \(2.00\) g of radon-220, what quantity will be left after \(1.00\) minute?

Because the rate is constant each second, this is an example of exponential decay, and we will use the formula \(a=a_0(1-r)^t\).

\(a_0 = 2.00\) g
\(r = 0.0133\)
\(t=60\) s

\[a=2.00(1-0.0133)^{60}\] \[a = 2.00(0.9867)^{60}\] \[a = 2.00(0.448)\] \[a = 0.896 \text{ g }\]

Compound Interest

Compound interest is interest earned on a sum of money over a certain time period that is then added to the original sum, so during the next time period there will be a larger sum to earn interest on. Each time period interest is earned on the original sum plus interest on the interest.

General Procedure

Because the rate of increase is constant for every time period, this is an example of exponential growth, so we can use the formula \(a=a_0(1+r)^t\) to calculate the effect of compound interest.


Suppose you deposit \(\$5,000\) in an account that adds \(1.6 \%\) interest every 3 months. What will be the value of the account in 7 years?

We will use:

\[a=a_0(1+r)^t\] \[a_0= \$5,000\] \[r = 1.6 \% = 0.016\]

Every three months is \(4\) times a year, and over \(7\) years, that’s 28 time periods.

\[t = 28\] \[a = \$5,000(1+0.016)^{28}\] \[a = \$5,000(1.016)^{28}\] \[a = \$5,000(1.56)\] \[a = \$7798\]


Depreciation is the term for the drop in value over the lifetime of things such as vehicles, homes, or pieces of manufacturing equipment as they age.

General Procedure

Accountants have different ways of calculating depreciation, but it can be treated as an exponential decay situation, using:



If the value of a certain car decreases at a rate of \(7 \%\) per year, what would it be worth after 4 years if it had a \(\$28,000\) purchase price?

Because the rate is constant each hour, this is an example of exponential decay, and we will use the formula:

\[a=a_0(1-r)^t\] \[a_0 = \$28,000\] \[r = 0.07\] \[t=4\] \[a=\$28,000(1-0.07)^4\] \[a = \$28,000(0.93)^4\] \[a = \$28,000(0.748)\] \[a = \$20,945\]

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