20.02 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, $h$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 (EXtended)

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 (extended)

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 (EXTENDED)

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 $f\left(x\right)=A\times a^{x-h}+k$f(x)=A×ax−h+k 

For the CORE course, you 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.  

 

Worked examples

Example 4

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