4.02 Transformations of exponential graphs

 

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 $y=3^x+1$y=3x+1.

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 $y=3^x+1$y=3x+1?

$y=3^x$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: Shifting the graph up four units 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

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-coordinates of the graph by a factor of three 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 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 3

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 4

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

 

 

 

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 5

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

 

 

 

Practice question

Question 4