<< Chapter < Page | Chapter >> Page > |
If we let $\text{\hspace{0.17em}}C=0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}D=0\text{\hspace{0.17em}}$ in the general form equations of the sine and cosine functions, we obtain the forms
The amplitude is $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ and the vertical height from the midline is $\text{\hspace{0.17em}}\left|A\right|.\text{\hspace{0.17em}}$ In addition, notice in the example that
What is the amplitude of the sinusoidal function $\text{\hspace{0.17em}}f(x)=\mathrm{-4}\mathrm{sin}(x)?\text{\hspace{0.17em}}$ Is the function stretched or compressed vertically?
Let’s begin by comparing the function to the simplified form $\text{\hspace{0.17em}}y=A\mathrm{sin}(Bx).$
In the given function, $\text{\hspace{0.17em}}A=\mathrm{-4},\text{\hspace{0.17em}}$ so the amplitude is $\text{\hspace{0.17em}}\left|A\right|=\left|\mathrm{-4}\right|=4.\text{\hspace{0.17em}}$ The function is stretched.
What is the amplitude of the sinusoidal function $\text{\hspace{0.17em}}f(x)=\frac{1}{2}\mathrm{sin}(x)?\text{\hspace{0.17em}}$ Is the function stretched or compressed vertically?
$\frac{1}{2}\text{\hspace{0.17em}}$ compressed
Now that we understand how $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ relate to the general form equation for the sine and cosine functions, we will explore the variables $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}D.\text{\hspace{0.17em}}$ Recall the general form:
The value $\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$ for a sinusoidal function is called the phase shift , or the horizontal displacement of the basic sine or cosine function . If $\text{\hspace{0.17em}}C>0,\text{\hspace{0.17em}}$ the graph shifts to the right. If $\text{\hspace{0.17em}}C<0,\text{\hspace{0.17em}}$ the graph shifts to the left. The greater the value of $\text{\hspace{0.17em}}\left|C\right|,\text{\hspace{0.17em}}$ the more the graph is shifted. [link] shows that the graph of $\text{\hspace{0.17em}}f(x)=\mathrm{sin}\left(x-\pi \right)\text{\hspace{0.17em}}$ shifts to the right by $\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$ units, which is more than we see in the graph of $\text{\hspace{0.17em}}f(x)=\mathrm{sin}\left(x-\frac{\pi}{4}\right),\text{\hspace{0.17em}}$ which shifts to the right by $\text{\hspace{0.17em}}\frac{\pi}{4}\text{\hspace{0.17em}}$ units.
While $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ relates to the horizontal shift, $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ indicates the vertical shift from the midline in the general formula for a sinusoidal function. See [link] . The function $\text{\hspace{0.17em}}y=\mathrm{cos}\left(x\right)+D\text{\hspace{0.17em}}$ has its midline at $\text{\hspace{0.17em}}y=D.$
Any value of $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ other than zero shifts the graph up or down. [link] compares $\text{\hspace{0.17em}}f(x)=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}f(x)=\mathrm{sin}\text{\hspace{0.17em}}x+2,\text{\hspace{0.17em}}$ which is shifted 2 units up on a graph.
Given an equation in the form $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{sin}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{cos}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$ $\frac{C}{B}\text{\hspace{0.17em}}$ is the phase shift and $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ is the vertical shift .
Determine the direction and magnitude of the phase shift for $\text{\hspace{0.17em}}f(x)=\mathrm{sin}\left(x+\frac{\pi}{6}\right)-2.$
Let’s begin by comparing the equation to the general form $\text{\hspace{0.17em}}y=A\mathrm{sin}(Bx-C)+D.$
In the given equation, notice that $\text{\hspace{0.17em}}B=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}C=-\frac{\pi}{6}.\text{\hspace{0.17em}}$ So the phase shift is
or $\text{\hspace{0.17em}}\frac{\pi}{6}\text{\hspace{0.17em}}$ units to the left.
Determine the direction and magnitude of the phase shift for $\text{\hspace{0.17em}}f(x)=3\mathrm{cos}\left(x-\frac{\pi}{2}\right).$
$\frac{\pi}{2};\text{\hspace{0.17em}}$ right
Determine the direction and magnitude of the vertical shift for $\text{\hspace{0.17em}}f(x)=\mathrm{cos}\left(x\right)-3.$
Let’s begin by comparing the equation to the general form $\text{\hspace{0.17em}}y=A\mathrm{cos}(Bx-C)+D.$
In the given equation, $\text{\hspace{0.17em}}D=\mathrm{-3}\text{\hspace{0.17em}}$ so the shift is 3 units downward.
Determine the direction and magnitude of the vertical shift for $\text{\hspace{0.17em}}f(x)=3\mathrm{sin}\left(x\right)+2.$
2 units up
Given a sinusoidal function in the form $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{sin}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$ identify the midline, amplitude, period, and phase shift.
Notification Switch
Would you like to follow the 'Precalculus' conversation and receive update notifications?