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A function that changes by a constant multiplier is called an exponential function. For example, these functions can be used to describe population changes in a country and how the value of a used car decreases over time.

Many functions containing a variable **exponent**, are called exponential functions. Formally, any function that can be written in the following form is an exponential function.

$y=a⋅b_{x}$

Therefore, for all exponential functions $y=a⋅b_{x},$ $a =0$ and $b>0,b =1.$

For an exponential function
$y=a⋅b_{x},$
$a$ represents the initial value and $b$ represents the constant multiplier. These values can be used to graph the function. Consider $y=10000⋅0.8_{x}.$
### 1

The initial value, $a,$ of an exponential function is the number without an exponent. In this case, $a=10000.$ The constant multiplier, $b$ is the number with the exponent. Here, $b=0.8.$

### 2

The initial value is the $y$-value when $x=0.$ It can also be thought of as the $y$-intercept of the function. Here, the initial value is $10000$ so $(0,10000)$ is a point on the graph.

### 3

### 4

Identify $a$ and $b$

Plot the initial value

Use the constant multiplier to find more points

When the $x$-value increases by $1,$ the $y$-value is multiplied by $b.$ Since $b=0.8,$ the $y$-value when $x=1$ is $10000⋅0.8=8000.$ Thus, $(1,8000)$ also lies on the graph of the function. Similarly, the point $(2,6400)$ lies on the graph because $8000⋅0.8=6400.$ These points are shown on the graph.

This process can be repeated until a general form of the graph emerges.

Draw the curve

Lastly, the graph can be drawn by connecting the points with a smooth curve.

Graph the exponential function using the function rule and describe its key features. $y=3⋅2_{x}$

Show Solution

The function $y=3⋅2_{x}$ has the initial value $a=3$ and the constant multiplier $b=2.$ Let us use these values to mark four points on the function's graph.

The function can now be graphed by connecting the points with a smooth curve.

- First the graph shows a $y$-intercept at $(0,3).$
- The function $y=3⋅2_{x}$ is greater than $0$ for all $x.$ Although the left-end of the graph approaches the $x$-axis it never intersects it. Thus, there is no $x$-intercept.
- As $x$ approaches $-∞$ the function continues to approach, but never becomes parallel with, the $x$-axis. Thus, the function increases for all $x.$

- Looking at the graph, we can see that the left end approaches $y=0$ and the right end extends upward. Thus, the end behavior of $y=3⋅2_{x}$ can be written as follows.

$Asx→-Asx→+ ∞,∞, y→0y→+∞ $ Let us show this in the graph.

In $1976,$ scientists discovered a rare population of Flemish Giant rabbits in a secluded forest. Since then, they've been monitoring the population. During the five years of the study, the number of rabbits could be modeled with the exponential function shown.

Use the graph to write the rule for the function, then interpret its initial value and constant multiplier.

Show Solution

To write an exponential function rule, we need the initial value of the function, $a,$ and the constant multiplier, $b.$ $y=a⋅b_{x}$ Notice that the graph starts at $(0,80).$ This means that $80$ is the initial value.

Since $a=80,$ we can write the following incomplete function rule. $y=80⋅b_{x}$ To determine $b,$ we can use another point on the graph.

The point $(1,100)$ lies on the graph. Thus, we can susbtitute $x=1$ and $y=100$ into the rule above and solve for $b.$$y=80⋅b_{x}$

SubstituteII

$x=1$, $y=100$

$100=80⋅b_{1}$

Solve for $b$

ExponentOne

$a_{1}=a$

$100=80⋅b$

RearrangeEqn

Rearrange equation

$80⋅b=100$

DivEqn

$LHS/80=RHS/80$

$b=80100 $

CalcQuot

Calculate quotient

$b=1.25$

The constant $r$ can then be interpreted as the *rate of growth*, in decimal form. A value of $0.06,$ for instance, means that the quantity increases by $6%$ over every unit of time. As is the case with all exponential functions, $a$ is the $y$-coordinate of the $y$-intercept.

The constant $r$ can then be interpreted as the *rate of decay*, in decimal form. A value of $0.12,$ for instance, would mean that the quantity decreases by $12%$ over every unit of time.

In an ideal environment, bacteria populations grow exponentially and can be modeled with an exponential growth function. The bacteria Lactobacillus acidophilus duplicates about once every hour. A single bacteria cell is placed in an ideal environment. State and interpret the constants $a$ and $r$ for the growth that will occur. Then, write a function rule describing this growth.

Show Solution

The constant $a$ is the initial value of the quantity, in this case the number of bacteria. There was only one bacteria placed in the environment, so $a$ is $1.$ The constant $r$ is the rate of growth. Since the bacteria duplicate every hour, the amount of bacteria doubles every hour. This corresponds to an increase by $100%.$ Thus, $r$ is $1.$ Substituting this into the rule of an exponential growth function gives $P(t)=1⋅(1+1)_{t},$ which can be simplified as $P(t)=2_{t}.$ Since this is an exponential growth function, population will grow faster and faster, without bound. In a real environment, this would not happen, since the available space and nutrition would have to be infinite. At some point, the environment would no longer be ideal, so the growth would slow down or stop.

During a time period, the number of carps in a small lake can be modeled by the function $H(t)=800⋅0.88_{t},$ where $t$ is the time in years. State whether the function shows growth or a decay, and then find the rate of growth or decay, $r.$ Finally, graph the function.

Show Solution

To begin, let's analyze the given function rule. It's written in the form $y=a⋅b_{x},$ where $a$ is the initial value and $b$ is constant multiplier/growth factor. The constant multiplier, $0.88,$ is less than $1,$ so it is a decay factor. Therefore, the function shows decay. Since the decay factor is always equal to $1−r,$ we can write the equation. $0.88=1−r,$ which can be solved for $r.$

Thus, the rate of decay is $0.12,$ or $12%,$ per year. The initial value is $800,$ and the constant multiplier is $0.88.$ Using this information, we can graph the exponential decay function by plotting some points that lie on $H$ and connecting them with a smooth curve.

When money is deposited to a savings account, interest is accrued, often yearly. Different types of interest work in different ways. When the interest earned is then added to the original amount, future interest accrues for a larger amount. This is called compound interest. To calculate the balance on the account at a specific time, an exponential growth function can be used. When the interest is compounded yearly, the balance can be modeled with a function.

$y=P(1+r)_{t}$

In this context, $P$ stands for the *principal*, which is the initial amount of money, and $r$ is the interest rate in decimal form. If the interest is not compounded yearly, the function looks a little different.

$y=P(1+nr )_{nt}$

The constant $n$ is the number of times the interest is compounded per year, while $r$ is still the annual interest rate. For an account with the principal $$100$ and an annual interest of $15%$ compounded twice a year, the growth function is:

$B(t)=100(1+20.15 )_{2t}$

Notice that this function grows continuously, whereas, in reality, the account balance only increases at the times of compound. Graphing the function together with the actual balance of the account will highlight how it can be used in practice.

Every time the interest is compounded, in this case every half year, the value of $B$ is equal to the account balance. However, at all other times, it is not. To find, for instance, the account balance after $1.75$ years, $B(1.5)$ should then be evaluated, since that was the last time interest compounded.One savings account, with a principal of $$100,$ offers an annual interest rate of $15%$ compounded twice a year. Find the balance in the account after $5$ years. Another savings account with the same principal will have the same balance after $5$ years. However, the interest is compounded monthly. Find the interest rate of the second account.

Show Solution

First, we'll find the function rule describing the growth of the first account. It is given that $P=100,r=0.15,$ and $n=2.$ Substituting these values in the compound interest formula gives $B(t)=100(1+20.15 )_{2t}.$ Let's simplify this function before continuing.

$B(t)=100(1+20.15 )_{2t}$

CalcQuot

Calculate quotient

$B(t)=100(1+0.075)_{2t}$

AddTerms

Add terms

$B(t)=100⋅1.075_{2t}$

$B(t)=100⋅1.075_{2t}$

Substitute

$t=5$

$B(5)=100⋅1.075_{2⋅5}$

Multiply

Multiply

$B(5)=100⋅1.075_{10}$

UseCalc

Use a calculator

$B(5)=206.10315…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$B(5)≈206.10$

The account balance is $$206.10$ after $5$ years. Now we can consider the second account. We know the balance of both accounts is equal, at least when both just had their interest compounded. This means that $B(t)$ also describes the growth in the second account. However, since the interest in the second account accrues monthly, or $12$ times a year, $n=12.$ Thus, the exponent in the rule should be $12t.$ By using the equality $2t=1/6⋅12t$ and the power of a power property, we can rewrite $B(t)$ so that it's possible to find the monthly interest rate. Having the exponent $12t$ means that the base of the power is equal to the monthly growth factor, which we can then use to find the monthly interest rate.

$B(t)=100⋅1.075_{2t}$

Rewrite

Rewrite $2t$ as $1/6⋅12t$

$B(t)=100⋅1.075_{1/6⋅12t}$

ProdInExponent

$a_{m⋅n}=(a_{m})_{n}$

$B(t)=100⋅(1.075_{1/6})_{12t}$

UseCalc

Use a calculator

$B(t)=100⋅(1.01212…)_{12t}$

RoundDec

Round to ${\textstyle 3 \, \ifnumequal{3}{1}{\text{decimal}}{\text{decimals}}}$

$B(t)≈100⋅1.012_{12t}$

We find an approximate monthly growth factor $1.012,$ which corresponds to a rate of growth that is $0.012.$ Thus, the monthly interest rate is roughly $1.2%.$

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