We have the following functions:
[tex]f_{1}(x)=sin(x) \\ f_{2}(x)=cos(x) \\ f_{3}(x)=tan(x)[/tex]
So let's analyze how to transform these functions by using key features.
1. Including amplitude.
To include the amplitude in a sine or cosine functions we simply multiply the function by A, so:
[tex]f_{1}(x)=Asin(x) \ or \\ f_{2}(x)=Acos(x)[/tex]
The absolute value of A is the amplitude. that is:
[tex]Amplitude=\left | A \right |[/tex]
The tangent function have no amplitude because this function grows without a bound.
2. Including period.
Let [tex]\omega[/tex] be a positive real number. The period of the sine or cosine functions is obtained as follows:
[tex]f_{1}(x)=sin(\omega x) \ and \ f_{2}(x)=cos(\omega x)[/tex]
The period T is given by:
[tex]T= \frac{2\pi}{\omega} \\ \\ \therefore \omega= \frac{2\pi}{T}[/tex]
Thus:
[tex]f_{1}(x)=sin(\frac{2\pi x}{T}) \ and \ f_{2}(x)=cos(\frac{2\pi x}{T})[/tex]
Regarding the tangent function:
[tex]f_{3}(x)=tan(\omega x)[/tex]
where the Period T is given by:
[tex]T= \frac{\pi}{\omega} \\ \\ \therefore \omega= \frac{\pi}{T}[/tex]
Thus:
[tex]f_{3}(x)=tan(\frac{\pi x}{T})[/tex]
3. Phase shift
The constant [tex]\phi[/tex] in the equations:
[tex]f_{1}(x)=sin(\omega x\pm \phi) \\ f_{2}(x)=cos(\omega x\pm \phi) \\ f_{3}(x)=tan(\omega x\pm \phi) [/tex]
creates a horizontal translation (shift) of the basic functions. To the left (if positive) or to the right (if negative). The number:
[tex]\frac{\phi}{\omega}[/tex]
is the phase shift.
4. Vertical shift
The constant [tex]B[/tex] in the equations:
[tex]f_{1}(x)=sin(x)+B \\ f_{2}(x)=cos(x)+B \\ f_{3}(x)=tan(x)+B[/tex]
creates a vertical translation (shift) of the basic functions. Upward (if positive) or downward (if negative)
5. Minimum point.
To find the minimum point in a sine or cosine functions let's take the functions:
[tex]f_{1}(x)=Asin(x) \ or \\ f_{2}(x)=Acos(x)[/tex]
The minimum point in a sine function is given when this function is minimum, that is, when [tex]x=\frac{3\pi}{2}[/tex]. On the other hand, the minimum point of the cosine function is given when [tex]x=\pi[/tex], then the minimum points are:
[tex]Sine \ function: \\ \\ min(\frac{3\pi}{2},-A) \\ \\ \\ Cosine \ function: \\ \\ min(\pi,-A)[/tex]
There is no a minimum point of the tangent function because it grows without a bound.
6. Maximum point.
To find the Maximum point in a sine or cosine functions let's take the functions:
[tex]f_{1}(x)=Asin(x) \ or \\ f_{2}(x)=Acos(x)[/tex]
So the Maximum point of the sine function is given when this function is Maximum, that is, when [tex]x=\frac{\pi}{2}[/tex]. On the other hand, the maximum point of the cosine function is given when [tex]x=0[/tex], then the maximum points are:
[tex]Sine \ function: \\ \\ Max(\frac{\pi}{2},A) \\ \\ \\ Cosine \ function: \\ \\ Max(0,A)[/tex]
There is no a maximum point in a tangent function because it grows negatively without a bound.
7. Example for the sine function.
As shown in Figure 1 we have the graph of the following function:
[tex]f_{1}(x)=3sin(\frac{x}{2}-\frac{\pi}{3})+4[/tex]
So the key features are:
[tex]Amplitude: \boxed{3} \\ \\ Period:T= \frac{2\pi}{\frac{1}{2}}=\boxed{4\pi} \\ \\ Phase \ shift:\boxed{\frac{\phi}{\omega}=\frac{2\pi}{3}} \\ \\ Vertical \ shift:\boxed{4} \\ \\ Maximum \ point: when \ \frac{x}{2}- \frac{\pi}{3}= \frac{\pi}{2} \therefore x= \frac{5\pi}{3} \ then \ \boxed{Max(\frac{5\pi}{3},7)} \\ \\ Minimum \ point:when \ \frac{x}{2}- \frac{\pi}{3}= \frac{3\pi}{2} \therefore x= \frac{11\pi}{3} \ then \ \boxed{min(\frac{11\pi}{3},1)}[/tex]
8. Example for the cosine function.
As shown in Figure 2 we have the graph of the following function:
[tex]f_{2}(x)=5cos(\frac{x}{2}+\frac{\pi}{3})+2[/tex]
So the key features are:
[tex]Amplitude: \boxed{5} \\ \\ Period:T= \frac{2\pi}{\frac{1}{2}}=\boxed{4\pi} \\ \\ Phase \ shift:\boxed{ \frac{\phi}{\omega}=\frac{2\pi}{3}} \\ \\ Vertical \ shift:\boxed{2} \\ \\ Maximum \ point: when \ \frac{x}{2}+\frac{\pi}{3}=0 \therefore x= \frac{-2\pi}{3} \ then \ \boxed{Max(-\frac{2\pi}{3},7)} \\ \\ Minimum \ point:when \ \frac{x}{2}+ \frac{\pi}{3}= -\pi \therefore x=-\frac{8\pi}{3} \\ or \ x=-\frac{8\pi}{3}+4\pi=\frac{4\pi}{3} \then \ \boxed{min(\frac{4\pi}{3},-3)}[/tex]
9. Example for the tangent function.
As shown in Figure 3 we have the graph of the following function:
[tex]f_{2}(x)=5tan(\frac{\pi x}{2}+\frac{\pi}{3})+2[/tex]
So the key features are:
[tex]Amplitude: DNE \\ \\ Period:T= \frac{\pi}{\frac{1}{2}}=\boxed{2\pi} \\ \\ Phase \ shift:\boxed{\frac{\phi}{\omega}=\frac{2\pi}{3}} \\ \\ Vertical \ shift:\boxed{2} \\ \\ Maximum \ point: DNE \\ \\ Minimum \ point:DNE[/tex]
