To obtain the graph of $y=Af\left(x-h\right)+k$y=Af(x−h)+k from the graph of $y=f\left(x\right)$y=f(x):
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:
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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.
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.
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.
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).
Of the two functions $y=2^x$y=2x and $y=4\times2^x$y=4×2x, which is increasing more rapidly for $x>0$x>0?
$y=2^x$y=2x
$y=4\times2^x$y=4×2x
Answer the following.
Determine the $y$y-intercept of $y=2^x$y=2x.
Hence or otherwise determine the $y$y-intercept of $y=2^x-2$y=2x−2.
Determine the horizontal asymptote of $y=2^x$y=2x.
Hence or otherwise determine the horizontal asymptote of $y=2^x-2$y=2x−2.
Consider the graph of $y=-3^x$y=−3x.
State the equation of the asymptote of $y=-3^x$y=−3x.
What would be the asymptote of $y=2-3^x$y=2−3x?
How many $x$x-intercepts would $y=2-3^x$y=2−3x have?
What is the domain of $y=2-3^x$y=2−3x?
$x<3$x<3
$x<2$x<2
$x>2$x>2
All real $x$x
What is the range of $y=2-3^x$y=2−3x?
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.
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
The graph represents an exponential function that has the form $y=a^x$y=ax.
State the equation of the function.