Recall from our lesson on transformation of functions, we summarised the impact of the parameters $a$a, $b$b, $c$c and $d$d, have on the graph of $y=af\left(b\left(x-c\right)\right)+d$y=af(b(x−c))+d, as follows:
To obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$y=af(b(x−c))+d from the graph of $y=f\left(x\right)$y=f(x):
- $a$a dilates(stretches) the graph by a factor of $a$a from the $x$x-axis
- When $a<0$a<0 the graph was reflected about the $x$x-axis
- $b$b dilates(stretches) the graph by a factor of $\frac{1}{b}$1b from the $y$y-axis
- When $b<0$b<0 the graph was reflected about the $y$y-axis
- $c$c translates the graph $c$c units horizontally
- $d$d translates the graph $d$d units vertically
Let's look at these more closely in relation to the graphs $f\left(x\right)=\sin\left(x\right)$f(x)=sin(x) and $f\left(x\right)=\cos\left(x\right)$f(x)=cos(x)and the impact on key features of these graphs. Use the applet below to observe the impact of $a$a, $b$b, $c$c and $d$d on $f\left(x\right)=a\sin\left(b\left(x-c\right)\right)+d$f(x)=asin(b(x−c))+d and $f\left(x\right)=a\cos\left(b\left(x-c\right)\right)+d$f(x)=acos(b(x−c))+d:
We can see the parameters had the expected impact on the graph and we can describe key features in relation to these.
Key features of the graph of $y=a\sin\left(b\left(x-c\right)\right)+d$y=asin(b(x−c))+dor $y=a\cos\left(b\left(x-c\right)\right)+d$y=acos(b(x−c))+d are:
- Principal axis: This is line line about which the graph oscillates. For the base graph of $y=\sin x$y=sinxor $y=\cos x$y=cosx this is the $x$x-axis(the line $y=0$y=0). Since the graph has been translated vertically by $d$d units, the principal axis has the equation $y=d$y=d.
- Amplitude: The vertical distance from the principal axis to the maximum point is the amplitude. Hence, as the base graph has an amplitude of $1$1 and graph has been dilated by a factor of $a$a the amplitude$=|a|$=|a|.
- Period: The period is the length of a cycle. This can be found as the horizontal distance between two successive maximums or minimums. As $b$b dilates the graph horizontally by a factor of $\frac{1}{b}$1b and the base graph has a period of $2\pi$2π, the transformed graph has period$=\frac{2\pi}{b}$=2πb.
- Phase shift: $c$c translates the graph $c$c units horizontally.
- Maximum: The maximum value of the function is $y=d+|a|$y=d+|a|
- Minimum: The minimum value of the function is $y=d-|a|$y=d−|a|
Worked example
Consider the graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4)+2 which are drawn below.
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| The graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4)+2 |
Starting with the graph of $y=\sin x$y=sinx, we can work through a series of transformations so that it coincides with the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4)+2.
We can first reflect the graph of $y=\sin x$y=sinx about the $x$x-axis. This is represented by applying a negative sign to the function (multiplying the function by $-1$−1).
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| The graph of $y=-\sin x$y=−sinx |
Then we can increase the amplitude of the function to match. This is represented by multiplying the $y$y-value of every point on $y=-\sin x$y=−sinx by $2$2.
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| The graph of $y=-2\sin x$y=−2sinx |
Next we can apply the period change that is the result of multiplying the $x$x-value inside the function by $3$3. This means that to get a particular $y$y-value, we can put in an $x$x-value that is $3$3 times smaller than before. The result is a horizontal dilation by a factor of $\frac{1}{3}$13, note the new graph will now cycle every $\frac{2\pi}{3}$2π3.
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| The graph of $y=-2\sin3x$y=−2sin3x |
Our next step will be to obtain the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4), and we can do so by applying a horizontal translation. In order to see what translation to apply, however, we first factorise the function into the form $y=-2\sin\left(3\left(x+\frac{\pi}{12}\right)\right)$y=−2sin(3(x+π12)).
In this form, we can see that the $x$x-values are increased by $\frac{\pi}{12}$π12 inside the function. This means that to get a particular $y$y-value, we can put in an $x$x-value that is $\frac{\pi}{12}$π12 smaller than before. Graphically, this corresponds to shifting the entire function to the left by $\frac{\pi}{12}$π12 units.
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| The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4) |
Lastly, we translate the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4) upwards by $2$2 units, to obtain the final graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4)+2.
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| The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4)+2 |
When we geometrically apply each transformation to the graph of $y=\sin x$y=sinx, it's important to consider the order of operations. If the function is written in the form $y=a\sin\left(b\left(x-c\right)\right)+d$y=asin(b(x−c))+d or $y=a\cos\left(b\left(x-c\right)\right)+d$y=acos(b(x−c))+d then the order is $b$b, $c$c, $a$a, $d$d.
Think: "horizontal dilate then translate, vertical dilate then translate".
Practice questions
Question 1
Consider the function $y=-3\cos x$y=−3cosx.
What is the maximum value of the function?
What is the minimum value of the function?
What is the amplitude of the function?
Select the two transformations that are required to turn the graph of $y=\cos x$y=cosx into the graph of $y=-3\cos x$y=−3cosx.
Vertical dilation.
AHorizontal translation.
BReflection across the $x$x-axis.
CVertical translation.
D
Question 2
Which combinations of transformations could be used to turn the graph of $y=\cos x$y=cosx into the graph of $y=-\cos x+3$y=−cosx+3?
Select the two correct options.
Reflection about the $x$x-axis, then translation $3$3 units down.
AReflection about the $x$x-axis, then translation $3$3 units up.
BTranslation $3$3 units up, then reflection about the $x$x-axis.
CTranslation $3$3 units down, then reflection about the $x$x-axis.
D
Question 3
Consider the given graph of $y=\cos\left(x+\frac{\pi}{2}\right)$y=cos(x+π2).
What is the amplitude of the function?
How can the graph of $y=\cos x$y=cosx be transformed into the graph of $y=\cos\left(x+\frac{\pi}{2}\right)$y=cos(x+π2)?
By reflecting it about the $x$x-axis, and then translating it horizontally $\frac{\pi}{2}$π2 units to the left.
ABy reflecting it about the $x$x-axis, and then translating it horizontally $\frac{\pi}{2}$π2 units to the right.
BBy translating it horizontally $\frac{\pi}{2}$π2 units to the right.
CBy changing the period of the function.
DBy translating it horizontally $\frac{\pi}{2}$π2 units to the left.
E
Question 4
Determine the equation of the graphed function given that it is of the form $y=\cos\left(x-c\right)$y=cos(x−c), where $c$c is the least positive value.
Question 5
Determine the equation of the graphed function given that it is of the form $y=\sin bx$y=sinbx or $y=\cos bx$y=cosbx, where $b$b is positive.