In general, we say that the graph of a function $y=f\left(x\right)$y=f(x) is vertically translated when the resulting graph is of the form $y=f\left(x\right)+c$y=f(x)+c, where $c$c is some non-zero constant.
Graphically speaking, a vertical translation takes each point on the graph of $y=f\left(x\right)$y=f(x) and adds (or subtracts) a constant to the $y$y-value of each point. For instance, we might have the graph of $y=\cos x$y=cosx as shown below, and all the points shift upwards according to the constant term.
$y=\cos x$y=cosx vertically translated upwards by a positive constant $c$c. |
The constant term does not necessarily have to be positive. In the case that $c$c is negative, that is $c<0$c<0, the graph of a function will vertically translate downwards.
$y=\sin x$y=sinx vertically translated downwards by a negative constant $c$c. |
For a negative value of $c$c, the graph of $y=f\left(x\right)$y=f(x) translates vertically downwards, although we would still write the equation of the resulting graph as $y=f\left(x\right)+c$y=f(x)+c.
Alternatively, we might instead write $y=f\left(x\right)-c$y=f(x)−c, where $c$c is redefined as the absolute value.
The graph of $y=\sin x+k$y=sinx+k has been vertically translated upwards by $5$5 units from $y=\sin x$y=sinx. What is the value of $k$k?
Think: The equation $y=\sin x+k$y=sinx+k is of the form $y=f\left(x\right)+c$y=f(x)+c where $c$c determines the direction and the magnitude of the vertical translation.
Do: A positive value of $k$k will translate the graph of $y=\sin x$y=sinx upwards. So $k=5$k=5.
Reflect: If instead we asked about the graph of $y=\sin x-k$y=sinx−k, how might the value of $k$k change?
Which of the following is the graph of $y=\sin x+3$y=sinx+3?
A graph of $y=\cos x$y=cosx is given.
Plot the resulting graph when $y=\cos x$y=cosx is translated $2$2 units down.
The function $y=\sin x$y=sinx is translated $2$2 units up.
Determine the equation of the new function after the translation.
The graphs of trigonometric functions like $y=\sin x$y=sinx and $y=\cos x$y=cosx have sections that are repeated in the $x$x-direction. The length of these repeated sections is called the wavelength, and we can understand a lot about the behaviour of the whole function by looking only at what happens within a single wavelength.
Typically, each trigonometric graph will have a minimum value, a maximum value, and and average value along the wavelength. The distance between the maximum (or minimum) value and the average value is called the amplitude.
If the range of the function is given by $\text{maximum value - minimum value }$maximum value - minimum value , the amplitude is then $\frac{\text{max - min }}{2}$max - min 2. Notice that the amplitude is always a positive number.
An equation of the form $y=a\sin x$y=asinx has an amplitude of $a$a units. If we start with the equation $y=\sin x$y=sinx, where $a=1$a=1, we can transform this equation by changing the amplitude.
Graphically this transformation corresponds to stretching the graph of $\sin x$sinx in the vertical direction when $a>1$a>1, and compressing the graph of $\sin x$sinx in the vertical direction when $00<a<1. We can also reflect the graph across the $x$x-axis by multiplying the whole function by $-1$−1.
This action of stretching and compressing the graph of a function is known as dilation. For trigonometric graphs like $\sin x$sinx and $\cos x$cosx, a vertical dilation by a factor of $a$a is equivalent to increasing the amplitude by a factor of $a$a.
State the amplitude of the function $f\left(x\right)=5\sin x$f(x)=5sinx.
Think: When we compare $f\left(x\right)$f(x) to the standard function $\sin x$sinx we can see that all the function values of $5\sin x$5sinx will be five times larger than all the function values of $\sin x$sinx. This means that the amplitude of $5\sin x$5sinx is also five times larger than the amplitude of $\sin x$sinx.
Do: The amplitude of $\sin x$sinx is $1$1, so the amplitude of $f\left(x\right)=5\sin x$f(x)=5sinx is $5\times1=5$5×1=5.
Reflect: We can obtain the graph of $f\left(x\right)=5\sin x$f(x)=5sinx by starting with the graph of $y=\sin x$y=sinx and applying a vertical dilation by a factor of $5$5.
The graph of $y=\cos x$y=cosx is reflected across the $x$x-axis, then compressed in the vertical direction so that its minimum value is $\frac{3}{4}$34. What is the equation of the resulting function. What is the amplitude?
Think: Recall that a reflection across the $x$x-axis corresponds to multiplying the function by $-1$−1. This will "switch" the location of the maximum and minimum values of the graph, but the amplitude will still be a positive value.
Do: Let's keep track of how the equation of the graph changes at each stage of the transformation.
$y=\cos x$y=cosx | $\rightarrow$→ | $y=-\cos x$y=−cosx | (Reflection across $x$x-axis) |
$y=-\cos x$y=−cosx | $\rightarrow$→ | $y=-\frac{3}{4}\cos x$y=−34cosx | (Vertical compression) |
The final equation of the resulting graph is $y=-\frac{3}{4}\cos x$y=−34cosx. The amplitude of this equation is $\frac{3}{4}$34.
Reflect: Compare the resulting equation with the original equation. The only difference is the constant multiple of $-\frac{3}{4}$−34. Notice that although this number is negative, the amplitude of the resulting equation is positive. In general, the amplitude of the equation $y=a\cos x$y=acosx is $\left|a\right|$|a|.
Consider the graph of the equation $y=\sin x$y=sinx. If this graph was vertically dilated by a factor of $9$9 from the $x$x-axis, what would be the equation of the resulting graph?
$y=9\sin x$y=9sinx
$y=\sin\frac{1}{9}x$y=sin19x
$y=\sin9x$y=sin9x
$y=\frac{1}{9}\sin x$y=19sinx
Determine the equation of the graphed function given that it is of the form $y=a\sin x$y=asinx or $y=a\cos x$y=acosx.
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.
Horizontal translation.
Reflection across the $x$x-axis.
Vertical translation.
We define the radian measure of an angle in terms of the length of the arc associated with the angle in the unit circle. There must be $2\pi$2π radians in a full circle because this is the length of the circumference. In the diagram above, the arc associated with the angle $\frac{2\pi}{3}$2π3 has length $\frac{2\pi}{3}$2π3.
We define the $\cos$cos and $\sin$sin functions as the horizontal and vertical coordinates of a point that moves on the unit circle. In the diagram above, we see that $\cos\frac{2\pi}{3}=-\frac{1}{2}$cos2π3=−12 and $\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$sin2π3=√32.
If we imagine the point moving counterclockwise on the unit circle so that the radius from the point makes an ever-increasing angle with the positive horizontal axis, eventually the angle exceeds $2\pi$2π; but the values of the $\cos$cos and $\sin$sin functions repeat the values from the angle $2\pi$2π smaller. We say $\sin$sin and $\cos$cos are periodic functions with period $2\pi$2π.
Thus, for any angle $x$x, there is a sequence of angles with the same value of $\sin x$sinx.
$...,x-4\pi,x-2\pi,x,x+2\pi,x+4\pi,x+6\pi,...$...,x−4π,x−2π,x,x+2π,x+4π,x+6π,...
Again, consider the angle $x$x made by the point moving around the unit circle. If a new angle $x'$x′ is defined by $x'=kx$x′=kx, we know that $\sin x'$sinx′ has period $2\pi$2π, but we see that $x'$x′ reaches $2\pi$2π when $x=\frac{2\pi}{k}$x=2πk. So, $\sin kx$sinkx and $\cos kx$coskx must have period $\frac{2\pi}{k}$2πk with respect to $x$x.
The function $\sin2x$sin2x begins to repeat when $2x=2\pi$2x=2π. That is, when $x=\pi$x=π. So, $\sin2x$sin2x has period $\pi$π. The period is multiplied by $\frac{1}{2}$12 when $x$x is multiplied by $2$2.
Thus, we see that for functions $\sin kx$sinkx and $\cos kx$coskx where $k$k is a constant, the period of the function with respect to $kx$kx is $2\pi$2π but the period with respect to $x$x is $\frac{2\pi}{k}$2πk.
We can use these ideas to deduce the formula for a sine or cosine function from a graph.
This graph looks like the graph of a cosine function since it has the value $1$1 at $0$0. However, the period is $3.2$3.2.
We know that $\cos kx$coskx has period $\frac{2\pi}{k}$2πk and, in this case, $\frac{2\pi}{k}=3.2$2πk=3.2. Therefore, $k=\frac{2\pi}{3.2}=\frac{2\pi}{\frac{16}{5}}=\frac{5\pi}{8}$k=2π3.2=2π165=5π8.
The graph must belong to the function given by $\cos\left(\frac{5\pi}{8}x\right)$cos(5π8x).
Consider the functions $f\left(x\right)=\sin x$f(x)=sinx and $g\left(x\right)=\sin3x$g(x)=sin3x.
State the period of $f\left(x\right)$f(x) in radians.
Complete the table of values for $g\left(x\right)$g(x).
$x$x | $0$0 | $\frac{\pi}{6}$π6 | $\frac{\pi}{3}$π3 | $\frac{\pi}{2}$π2 | $\frac{2\pi}{3}$2π3 | $\frac{5\pi}{6}$5π6 | $\pi$π | $\frac{7\pi}{6}$7π6 | $\frac{4\pi}{3}$4π3 |
---|---|---|---|---|---|---|---|---|---|
$g\left(x\right)$g(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
State the period of $g\left(x\right)$g(x) in radians.
What transformation of the graph of $f\left(x\right)$f(x) results in the graph of $g\left(x\right)$g(x)?
Vertical dilation by a factor of $\frac{1}{3}$13
Vertical dilation by a factor of $3$3
Horizontal dilation by a factor of $\frac{1}{3}$13
Horizontal dilation by a factor of $3$3
The graph of $f\left(x\right)$f(x) has been provided below.
By moving the points, graph $g\left(x\right)$g(x).
Consider the function $f\left(x\right)=\cos5x$f(x)=cos5x.
Determine the period of the function in radians.
What is the maximum value of the function?
What is the minimum value of the function?
Graph the function for $0\le x\le\frac{4}{5}\pi$0≤x≤45π.
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.
Phase shift for trigonometric functions means moving the graph of the function to the right or to the left. This transformation occurs when a constant is added to (or subtracted from) the angle or number to which the function is applied.
For example, the following functions include a phase shift transformation.
$\sin\left(\theta+\frac{\pi}{4}\right)$sin(θ+π4)
$\cos(x-0.5)$cos(x−0.5)
$\tan\left(\alpha+\frac{22}{7}\right)$tan(α+227)
The following graph shows the functions $\cos x$cosx and $\cos(x+0.4)$cos(x+0.4) on the same axes.
The graph of $\cos x$cosx is shown in black. It can be seen that the graph of $\cos(x+0.4)$cos(x+0.4) is the graph of $\cos x$cosx shifted to the left by the amount $0.4$0.4.
The dotted lines drawn on the diagram are intended to show that the function $\cos(x+0.4)$cos(x+0.4) when $x=0.5$x=0.5 attains the same value reached by $\cos x$cosx when $x=0.9$x=0.9. Thus, the shift is to the left.
The following graph looks like the graph of $\sin x$sinx with a phase shift of $1.05$1.05 to the right.
The graph must belong to the function given by $\sin(x-1.05)$sin(x−1.05). The phase shift to the right has been brought about by adding $-1.05$−1.05 to $x$x.
A more precisely drawn horizontal scale might reveal that the graph actually crosses the axis at $1.047$1.047 which is approximately $\frac{\pi}{3}$π3. So, another way of writing the function is with the expression $\sin\left(x-\frac{\pi}{3}\right)$sin(x−π3).
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.
By reflecting it about the $x$x-axis, and then translating it horizontally $\frac{\pi}{2}$π2 units to the right.
By translating it horizontally $\frac{\pi}{2}$π2 units to the right.
By changing the period of the function.
By translating it horizontally $\frac{\pi}{2}$π2 units to the left.
Consider the function $f\left(x\right)=\cos x$f(x)=cosx and $g\left(x\right)=\cos\left(x-\frac{\pi}{2}\right)$g(x)=cos(x−π2).
Complete the table of values for both functions.
$x$x | $0$0 | $\frac{\pi}{2}$π2 | $\pi$π | $\frac{3\pi}{2}$3π2 | $2\pi$2π |
---|---|---|---|---|---|
$f\left(x\right)$f(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
$g\left(x\right)$g(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Using the table of values, what transformation of the graph of $f\left(x\right)$f(x) results in the graph of $g\left(x\right)$g(x)?
vertical translation $\frac{\pi}{2}$π2 units downwards
horizontal translation $\frac{\pi}{2}$π2 units to the left
horizontal translation $\frac{\pi}{2}$π2 units to the right
vertical translation $\frac{\pi}{2}$π2 units upwards
The graph of $f\left(x\right)$f(x) has been provided below.
By moving the points, graph $g\left(x\right)$g(x).
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.