3.02 Exponential functions

Identify characteristics of exponential functions 

The mathematical function which models natural growth and decay phenomena is known as the exponential model. What distinguishes it from other functions you may have encountered is that the independent variable is in the exponent, (the exponent is also called the power or index). 

Examples include $y=2^x$y=2x, $y=3\times2^{-x}$y=3×2−x, $y=8\times\left(0.5\right)^x+1$y=8×(0.5)x+1, $y=4-5^x$y=4−5x, etc.

If we first consider the function $y=2^x$y=2x, and substitute the consecutive integer value of $x$x into it, starting from $x=-3$x=−3 and finishing at $x=3$x=3, we could develop a table of values for the function. 

So for $x=-3$x=−3, $y=2^{-3}=\frac{1}{2^3}=\frac{1}{8}$y=2−3=123​=18​. For $x=-2$x=−2, $y=\frac{1}{4}$y=14​ and continuing:

$x$x	$-3$−3	$-2$−2	$-1$−1	$0$0	$1$1	$2$2	$3$3
$y$y	$\frac{1}{8}$18​	$\frac{1}{4}$14​	$\frac{1}{2}$12​	$1$1	$2$2	$4$4	$8$8

Typical of exponential growth we see that initially there are small increases in $y$y , but as we continue to increase $x$x, the rate of increase in the size of $y$y is increasing. That is to say the $y$y values are getting larger faster as $x$x steadily increases. Populations, by and large, behave like this - the more people that live in a city, the faster the total population grows. Investments in bank accounts are similar - the more money you have in the bank, the more interest you earn. 

Things can work in reverse as well. Consider the function given by $y=120\times2^{-x}$y=120×2−x. This table shows what happens to $y$y for integer values between $0$0 and $6$6.

$x$x	$0$0	$1$1	$2$2	$3$3	$4$4	$5$5	$6$6
$y$y	$120$120	$60$60	$30$30	$15$15	$7.5$7.5	$3.75$3.75	$1.875$1.875

 

We can use the table to model depreciation. For example, imagine an expensive car, say $$120000$120000, that halves its value every $4$4 years. After the first four years ($x=1$x=1), it would be worth $$60000$60000. After the second $4$4 years ($x=2$x=2) it would be worth $$30000$30000. After $24$24 years ($x=6$x=6), the car's value is reduced to $$1875$1875.

Understanding the shape of the exponential graph. 

Below is the graph of $y=2^x$y=2x. Here are some key features worth noting:

• The function is increasing over its domain
• The rate of change increases as $x$x gets larger and larger
• It has a $y$y-intercept of $1$1 because,at $x=0$x=0, $y=2^0=1$y=20=1
• The point $\left(1,2\right)$(1,2) can help us graph the function
• End behavior
  • As the $x$x tends to $-\infty$−∞, the curve draws closer and closer to the $x$x-axis the further it is from the origin. The $x$x-axis becomes what is known as a horizontal asymptote - a line that a curve approaches without ever touching it.
  • As the $x$x tends to $\infty$∞, the curve moves upward away from the $x$x-axis. As it moves upward the bend becomes more pronounced as the rate of change in the y values becomes larger and larger. 

• The domain is all of the real numbers and the range is $y>0$y>0

 

 

The function $y=2^{-x}$y=2−x is graphed below. Here are some key features worth noting:

• The function is decreasing over its domain
• The rate of change approaches $0$0 as $x$x gets larger and larger
• It has a $y$y-intercept of $1$1 because,at $x=0$x=0, $y=2^{-0}=1$y=2−0=1
• The point $\left(-1,2\right)$(−1,2) can help us graph the function
• End behavior
  • As the $x$x tends to $-\infty$−∞, the curve moves upward away from the $x$x-axis. As it moves upward the bend becomes more pronounced as the rate of change in the $y$y values becomes steeper. 
  • As the $x$x tends to $\infty$∞, the curve draws closer and closer to the $x$x-axis the further it is from the origin. The $x$x-axis becomes what is known as a horizontal asymptote - a line that a curve approaches without ever touching it.
• The domain is all of the real numbers and the range is $y>0$y>0

 

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

 

 

Transformations of exponential functions

The exponential curve given by $y=a\times b^{n(x-h)}+k$y=a×bn(x−h)+k represents a transformation of the basic curve $y=b^x$y=bx.

Whilst the general form is a comprehensive tool for sketching exponential curves, there are a few simpler observations to keep in mind. We can summarize them using examples as shown in this table for the parent function $y=3^x$y=3x:

Specific Example	Observation
$y=-3^x$y=−3x	Reflect $y=3^x$y=3x across the $x$x-axis
$y=3^{x-5}$y=3x−5	Translate $y=3^x$y=3x horizontally  to the right by $5$5 units
$y=3^x-5$y=3x−5	Translate $y=3^x$y=3x vertically downward by $5$5 units
$y=2\times3^x$y=2×3x	Vertical stretch away from $x$x-axis, double every $y$y value of  $y=3^x$y=3x

More complex forms of the exponential require more thought. For example, the function $y=3^{2x-5}$y=32x−5 is quite interesting to think about. The applet below can produce the graph as a plot of points, but we can think about what the curve might look like without it.

For example, we can rewrite the function as follows:

 $y=3^{2\left(x-\frac{5}{2}\right)}=\left(3^2\right)^{\left(x-\frac{5}{2}\right)}=9^{\left(x-\frac{5}{2}\right)}.$y=32(x−52​)=(32)(x−52​)=9(x−52​).

Hence, the function could be thought of as the function $y=9^x$y=9x translated to the right by $2\frac{1}{2}$212​ units.

 

Exploration

The applet below is extremely versatile, but we need to keep in mind that it is a learning tool exploring the effects of the different constants involved.

Try to create the four graphs shown in the table below and note the key features of the graphs.

Function	$y$y-intercept	Asymptote	Increasing/decreasing
$y=2^x$y=2x
$y=3^{-x+1}$y=3−x+1
$y=3\times4^x-2$y=3×4x−2
$y=\left(0.5\right)^x$y=(0.5)x

After experimenting with these, try other combinations of constants. What can you learn? 

 

Bases between 0 and 1

One final point that should be noted is that a curve like $y=\left(0.5\right)^x$y=(0.5)x is none other than $y=2^{-x}$y=2−x in disguise. Thus: 

$y=\left(0.5\right)^x=\left(\frac{1}{2}\right)^x=\frac{1}{2^x}=2^{-x}$y=(0.5)x=(12​)x=12x​=2−x 

 

In a similar way we can say that $y=\left(\frac{1}{b}\right)^x=b^{-x}$y=(1b​)x=b−x, and so every exponential curve of the form $y=b^x$y=bx, with a base $b$b in the interval $00<b<1, can be re-expressed as $y=\left(\frac{1}{b}\right)^{-x}$y=(1b​)−x. Since $b$b is a positive number, this means that exponential functions of the form  $y=b^x$y=bx where $00<b<1 are in fact decreasing curves.  

 

Practice questions

Question 4

Question 5

 

Finding and using an exponential equation

Under certain circumstances, we are able to deduce a function's equation from a few known points the associated curve passes through. For example it may be possible when the form of the equation is known.

Worked examples

question 6

Consider the curve of a function with the exponential form $y=a\left(b^x\right)$y=a(bx) which passes through $\left(2,18\right)$(2,18) and $\left(5,486\right)$(5,486). Find the equation of the curve.

Think: These points must satisfy the equation (that is, when substituted, will make the equation true). Thus we know that:

$18$18	$=$=	$ab^2$ab2     (1)
$486$486	$=$=	$ab^5$ab5     (2)

Do:

Dividing equation (2) by equation (1) gives:

$\frac{486}{18}$48618​	$=$=	$\frac{ab^5}{ab^2}$ab5ab2​	Performing the division
$27$27	$=$=	$b^3$b3	Simplifying (assume $a\ne0$a≠0)
$3$3	$=$=	$b$b	Take the cube root of both sides

 

Thus since $ab^2=18$ab2=18, we have $9a=18$9a=18 and $a=2$a=2. So we identify the function as given by $y=2\left(3\right)^x$y=2(3)x.

Reflect: Could you have solved this using a different strategy?

 

Question 7

Fill in the table for values of a function of the form $y=b^{-x}+c$y=b−x+c as shown here:

$x$x	$-2$−2	$-1$−1	$0$0	$1$1	$2$2
$y$y		$5$5		$3.5$3.5

Think: We know the two points $\left(-1,5\right)$(−1,5) and $\left(1,3.5\right)$(1,3.5), so we form two equations as before:

$5$5	$=$=	$b^1+c$b1+c       $(1)$(1)
$3.5$3.5	$=$=	$b^{-1}+c$b−1+c     $(2)$(2)

Do: This means that from equation (2) we have $c=3.5-\frac{1}{b}$c=3.5−1b​ and so substituting this into equation (1) we have that $5=b+\left(3.5-\frac{1}{b}\right)$5=b+(3.5−1b​) or when simplified $1.5=b-\frac{1}{b}$1.5=b−1b​.

Solving for $b$b we have:

$1.5$1.5	$=$=	$b-\frac{1}{b}$b−1b​
$1.5b$1.5b	$=$=	$b^2-1$b2−1
$3b$3b	$=$=	$2b^2-2$2b2−2
$2b^2-3b-2$2b2−3b−2	$=$=	$0$0
$\left(2b+1\right)\left(b-2\right)$(2b+1)(b−2)	$=$=	$0$0
$b$b	$=$=	$-\frac{1}{2},2$−12​,2

 So there seem to be two possibilities. By substituting back into equation (1) above we find that if $b=-\frac{1}{2}$b=−12​, we find $c=5\frac{1}{2}$c=512​, and for $b=2$b=2 we find $c=3$c=3.

However, the first pairing, delivering $y=\left(-\frac{1}{2}\right)^{-x}+5\frac{1}{2}$y=(−12​)−x+512​ needs to be discarded, because the base b is defined to be a positive number.

So the correct solution becomes $y=2^{-x}+3$y=2−x+3.

$x$x	$-2$−2	$-1$−1	$0$0	$1$1	$2$2
$y$y	$7$7	$5$5	$4$4	$3.5$3.5	$3.25$3.25

 

 

 

Practice questions

Question 8

Question 9

Question 10