7.05 Graphs and transformations of exponential functions

Recall from our lesson 3.05 on transformations of functions, we summarised the impact of the parameters $A$A, $b$b, $h$h and $k$k, have on the graph of $y=Af\left(b\left(x-h\right)\right)+k$y=Af(b(x−h))+k, as follows:

Let's look at these more closely in relation to the graphs $f(x)=a^x$f(x)=ax and $f\left(x\right)=A\times a^{b\left(x-h\right)}+k$f(x)=A×ab(x−h)+k and the impact the parameters have on the key features. Use the applet below to observe the impact of $A$A, $b$b, $h$h and $k$k for a particular $a$a value:

Created with Geogebra

Did the parameters have the expected effect?

We can see in particular, the vertical translation by $k$k units causes the horizontal asymptote to become $y=k$y=k.

 

Worked examples

Example 1

Sketch the graph of $y=3^x+4$y=3x+4.

Think: What does the base graph of $y=3^x$y=3x look like? And what transformations would take $y=3^x$y=3x to $y=3^x+4$y=3x+4?

$y=3^x$y=3x is an increasing function with $y$y-intercept $\left(0,1\right)$(0,1), goes through the point $\left(1,3\right)$(1,3) and has a horizontal asymptote $y=0$y=0. To transform this graph we need to translate the graph $4$4 units up.

Do: Translating 4 unit up (adding $4$4 to each $y$y value), the points become $\left(0,5\right)$(0,5) and $\left(1,7\right)$(1,7) and the horizontal asymptote is shifted to $y=4$y=4.

Sketch a dotted line for the asymptote, plot our two points $\left(0,5\right)$(0,5) and $\left(1,7\right)$(1,7) (you can plot more points to obtain a more accurate sketch or to give you confidence in the shape of the graph). Draw a smooth increasing curve through these points and approaching the asymptote to the left.

 

Example 2

(a) Sketch the graph of $y=2^{x+3}$y=2x+3.

Think: What does the base graph of $y=2^x$y=2x look like? And what transformations would take $y=2^x$y=2x to $y=2^{x+3}$y=2x+3?

$y=2^x$y=2x is an increasing function with $y$y-intercept $\left(0,1\right)$(0,1), goes through the point $\left(1,2\right)$(1,2) and has a horizontal asymptote $y=0$y=0. To transform this graph we need to translate the graph $3$3 units to the left.

Do: Translating $3$3 units left (subtracting $3$3 from every $x$x value), the points become $\left(-3,1\right)$(−3,1) and $\left(-2,2\right)$(−2,2) and the horizontal asymptote remains at $y=0$y=0.

Sketch a dotted line for the asymptote, plot our two points $\left(-3,1\right)$(−3,1) and $\left(-2,2\right)$(−2,2). Let's also find the $y$y-intercept: when $x=0$x=0, $y=2^{\left(0+3\right)}=8$y=2(0+3)=8, thus the $y$y-intercept is $\left(0,8\right)$(0,8). Each point will move three units to the left, additional points can be sketched to have a more accurate shape for the graph. For example, $\left(4,16\right)$(4,16) becomes $\left(1,16\right)$(1,16). Draw a smooth increasing curve through these points and approaching the asymptote to the left.

(b) Using index laws rewrite the function in the form $y=A\times2^x$y=A×2x.

Think: Recall our index law $a^{m+n}=a^m\times a^n$am+n=am×an and use it here to split the term $2^{x+3}$2x+3.

Do:

$y$y	$=$=	$2^{x+3}$2x+3
	$=$=	$2^x\times2^3$2x×23
	$=$=	$8\times2^x$8×2x

Reflect: This means that the horizontal translation in this example is equivalent to a vertical dilation by a factor of $8$8. Look back at the graph and confirm this. If required we could always use our index laws to rewrite a horizontal translation as a vertical dilation.

 

Example 3

Sketch the graph of $y=3\times2^x+1$y=3×2x+1.

Think: What does the base graph of $y=2^x$y=2x look like? And what transformations would take $y=2^x$y=2x to $y=3\times2^x+1$y=3×2x+1?

$y=2^x$y=2x is an increasing function with $y$y-intercept $\left(0,1\right)$(0,1), goes through the point $\left(1,2\right)$(1,2) and has a horizontal asymptote $y=0$y=0. To transform this graph we need to dilate the graph vertically by a factor of three and then translate the graph $1$1unit up.

Do: Stretching the $y$y-coordinates of the graph by a factor of three (multiply $y$y values by $3$3) we would have the points $\left(0,3\right)$(0,3) and $\left(1,6\right)$(1,6). Then shifting the graph up one unit (add $1$1 to $y$y values) the points become $\left(0,4\right)$(0,4) and $\left(1,7\right)$(1,7) and the horizontal asymptote is shifted to $y=1$y=1.

Sketch a dotted line for the asymptote, plot our two points $\left(0,4\right)$(0,4) and $\left(1,7\right)$(1,7) (you can plot more points to obtain a more accurate sketch or to give you confidence in the shape of the graph). Draw a smooth increasing curve through these points and approaching the asymptote to the left.

 

Example 4

Sketch the graph of $y=-2\times3^{x+2}$y=−2×3x+2.

Think: What does the base graph of $y=3^x$y=3x look like? And what transformations would take $y=3^x$y=3x to $y=-2\times3^{x+2}$y=−2×3x+2?

$y=3^x$y=3x is an increasing function with $y$y-intercept $\left(0,1\right)$(0,1), goes through the point $\left(1,3\right)$(1,3) and has a horizontal asymptote $y=0$y=0. To transform this graph we need to dilate the graph vertically by a factor of two, reflect the graph across the $x$x-axis and then translate the graph $2$2 units to the left.

Do: Stretching the $y$y-coordinates of the graph by a factor of two and reflecting (multiply $y$y values by $-2$−2) we would obtain the points $\left(0,-2\right)$(0,−2) and $\left(1,-6\right)$(1,−6). Then shifting the graph left two units (subtract $2$2 from $x$x values) the points become $\left(-2,-2\right)$(−2,−2) and $\left(-1,-6\right)$(−1,−6) and the horizontal asymptote remains at $y=0$y=0.

Sketch a dotted line for the asymptote, plot our two points $\left(-2,-2\right)$(−2,−2) and $\left(-1,-6\right)$(−1,−6). Let's also find the $y$y-intercept: when $x=0$x=0, $y=-2\times3^2=-18$y=−2×32=−18, thus the $y$y-intercept is $\left(0,-18\right)$(0,−18). Draw a smooth decreasing curve through these points and approaching the asymptote to the left.

 

Example 5

Sketch the graph of $y=2^{-3x}+4$y=2−3x+4.

Think: When dealing with a $b$b value that is not $1$1 or $-1$−1 we can approach the situation in either of the following ways:

• We could consider this a horizontal dilation of the base graph $y=2^{-x}$y=2−x by a factor of $\frac{1}{3}$13​
• Or we could use our index laws to rewrite $y=2^{-3x}$y=2−3x as $y=\left(2^3\right)^{-x}=8^{-x}$y=(23)−x=8−x, and then consider transformations on the base graph of $y=8^{-x}$y=8−x

We can in fact always use index laws to remove the need to deal with horizontal dilations or translations.

Let's use our second option here and consider our base graph $y=8^{-x}$y=8−x. This is a decreasing function with $y$y-intercept $\left(0,1\right)$(0,1), goes through the point $\left(-1,8\right)$(−1,8) and has a horizontal asymptote $y=0$y=0. To transform this graph we need to translate the graph $4$4 units up.

Do: Translating the points and asymptote we obtain $\left(0,5\right)$(0,5) and $\left(-1,12\right)$(−1,12) and horizontal asymptote of $y=4$y=4.

Sketch a dotted line for the asymptote, plot our two points $\left(0,5\right)$(0,5) and $\left(-1,12\right)$(−1,12). Draw a smooth decreasing curve through these points and approaching the asymptote to the right.

Note: The examples above did not contain an $x$x-intercept but a graph such as $y=3\times2^x-12$y=3×2x−12 would. If required to label an $x$x-intercept we can solve for $y=0$y=0 in some cases. For instance, if we can rearrange the equation such that both sides are written with the same base. We will look more thoroughly at solving these types of equations in the last lesson of this chapter. To solve more complex cases for the time being we will use technology to help us find the $x$x-intercepts.

 

Example 6

Find the $x$x-intercept of $y=3\times2^x-12$y=3×2x−12.

Think: Let $y=0$y=0 and rearrange the equation to $2^x=...$2x=.... Then check if both sides of the equation can be written as a power of $2$2.

Do:

$3\times2^x-12$3×2x−12	$=$=	$0$0
$3\times2^x$3×2x	$=$=	$12$12
$2^x$2x	$=$=	$4$4

Then since both sides can be written as a power of two we can equate the powers:

$2^x$2x	$=$=	$2^2$22
Therefore, $x$x	$=$=	$2$2

So the $x$x-intercept is $\left(2,0\right)$(2,0).

 

Practice questions

Question 1

Question 2

Question 3

Question 4

 

 

Function from a graph

In the previous examples, we have seen how an exponential function can be used to produce a sketch.  We will also want to determine the rule for an exponential function from a graph.

The general rule for an exponential function is $f\left(x\right)=A\times a^{b\left(x-h\right)}+k$f(x)=A×ab(x−h)+k 

For this course, we only need to consider exponential functions of the form $f\left(x\right)=a^{x-h}+k$f(x)=ax−h+k where $a>0$a>0.  

So we may need to determine the values of each of the parameters $a$a, $h$h, and $k$k.  

 

Worked examples

Example 7

Determine the rule for the exponential function of the form $f(x)=a^x+k$f(x)=ax+k, represented by the graph below:

Think: This graph represents exponential decay so the base parameter, $a$a, for the exponential function will be between $0$0 and $1$1; 

The horizontal asymptote is not on the $x$x-axis so there is a vertical translation, determined by the value of $k$k.  

Do: The horizontal asymptote is given by $y=1$y=1 so $k=1$k=1.

We can now determine the base parameter, $a$a, substituting the given point: $x=1$x=1, $y=1.6$y=1.6

From the graph, we can see that $f(1)=1.6$f(1)=1.6. 

$f(1)$f(1)	$=$=	$1.6$1.6
$a^1+1$a1+1	$=$=	$1.6$1.6
$a$a	$=$=	$0.6$0.6

Therefore, the rule for this function is:

$f(x)=0.6^x+1$f(x)=0.6x+1

 

EXAMPLE 8

The graph below represents an exponential function with the form $f(x)=2^{x-h}+k$f(x)=2x−h+k.  Determine the values of the parameters $h$h and $k$k and determine the rule for this function.

Think: The graph passes through the point $\left(0,1\right)$(0,1) but the horizontal asymptote is not on the $x$x-axis so there is a vertical translation produced by parameter $k\ne0$k≠0;

It will be easiest to determine the vertical translation first. Then use the $y$y-intercept to determine the other parameter.

Do: The horizontal asymptote is given by $y=-1$y=−1 so $k=-1$k=−1.

We can now determine the final parameter from the vertical intercept at $0,1$0,1.

$f(0)$f(0)	$=$=	$1$1
$2^{0-h}-1$20−h−1	$=$=	$1$1
$2^{-h}$2−h	$=$=	$2$2
$\therefore\ -h$∴ −h	$=$=	$1$1	Equating indices.
$h$h	$=$=	$-1$−1

Therefore, $f(x)=2^{x+1}-1$f(x)=2x+1−1.

Reflect: When the function value ($y$y-coordinate) is equal to $1+k$1+k, the $x$x-coordinate gives the value of $h$h. This can give us a second way to find or check the value of $h$h.

 

Practice question

Question 5

Question 6