Mathematics, 13.12.2019 09:31 nothingworksoutforme

Describe how to transform a trigonometric function from the standard trig function f(x) = sin x, f(x) = cos x, or f(x) = tan x using key features. include amplitude, period, phase shift, vertical shift, the minimum point and a maximum point.

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Answer from: ctdavis

We have the following functions:

$f_{1}(x)=sin(x) \\ f_{2}(x)=cos(x) \\ f_{3}(x)=tan(x)$

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:

$f_{1}(x)=Asin(x) \ or \\ f_{2}(x)=Acos(x)$

The absolute value of A is the amplitude. that is:

$Amplitude=\left | A \right |$

The tangent function have no amplitude because this function grows without a bound.

2. Including period.

Let $\omega$ be a positive real number. The period of the sine or cosine functions is obtained as follows:

$f_{1}(x)=sin(\omega x) \ and \ f_{2}(x)=cos(\omega x)$

The period T is given by:

$T= \frac{2\pi}{\omega} \\ \\ \therefore \omega= \frac{2\pi}{T}$

Thus:

$f_{1}(x)=sin(\frac{2\pi x}{T}) \ and \ f_{2}(x)=cos(\frac{2\pi x}{T})$

Regarding the tangent function:

$f_{3}(x)=tan(\omega x)$

where the Period T is given by:

$T= \frac{\pi}{\omega} \\ \\ \therefore \omega= \frac{\pi}{T}$

Thus:

$f_{3}(x)=tan(\frac{\pi x}{T})$

3. Phase shift

The constant $\phi$ in the equations:

$f_{1}(x)=sin(\omega x\pm \phi) \\ f_{2}(x)=cos(\omega x\pm \phi) \\ f_{3}(x)=tan(\omega x\pm \phi)$

creates a horizontal translation (shift) of the basic functions. To the left (if positive) or to the right (if negative). The number:

$\frac{\phi}{\omega}$

is the phase shift.

4. Vertical shift

The constant

in the equations:

$f_{1}(x)=sin(x)+B \\ f_{2}(x)=cos(x)+B \\ f_{3}(x)=tan(x)+B$

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:

$f_{1}(x)=Asin(x) \ or \\ f_{2}(x)=Acos(x)$

The minimum point in a sine function is given when this function is minimum, that is, when $x=\frac{3\pi}{2}$ . On the other hand, the minimum point of the cosine function is given when $x=\pi$ , then the minimum points are:

$Sine \ function: \\ \\ min(\frac{3\pi}{2},-A) \\ \\ \\ Cosine \ function: \\ \\ min(\pi,-A)$

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:

$f_{1}(x)=Asin(x) \ or \\ f_{2}(x)=Acos(x)$

So the Maximum point of the sine function is given when this function is Maximum, that is, when $x=\frac{\pi}{2}$ . On the other hand, the maximum point of the cosine function is given when x=0

, then the maximum points are:

$Sine \ function: \\ \\ Max(\frac{\pi}{2},A) \\ \\ \\ Cosine \ function: \\ \\ Max(0,A)$

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:

$f_{1}(x)=3sin(\frac{x}{2}-\frac{\pi}{3})+4$

So the key features are:

$Amplitude: \boxed{3} \\ \\ Period:T= \frac{2\pi}{\frac{1}{2}}=\boxed{4\pi} \\ \\ Phase \ shift:\boxed{\frac{\phi}{\omega}=\frac{2\pi}{3}} \\ \\ Vertical \ shift:\boxed{4} \\ \\ Maximum \ point: when \ \frac{x}{2}- \frac{\pi}{3}= \frac{\pi}{2} \therefore x= \frac{5\pi}{3} \ then \ \boxed{Max(\frac{5\pi}{3},7)} \\ \\ Minimum \ point:when \ \frac{x}{2}- \frac{\pi}{3}= \frac{3\pi}{2} \therefore x= \frac{11\pi}{3} \ then \ \boxed{min(\frac{11\pi}{3},1)}$

8. Example for the cosine function.

As shown in Figure 2 we have the graph of the following function:

$f_{2}(x)=5cos(\frac{x}{2}+\frac{\pi}{3})+2$

So the key features are:

$Amplitude: \boxed{5} \\ \\ Period:T= \frac{2\pi}{\frac{1}{2}}=\boxed{4\pi} \\ \\ Phase \ shift:\boxed{ \frac{\phi}{\omega}=\frac{2\pi}{3}} \\ \\ Vertical \ shift:\boxed{2} \\ \\ Maximum \ point: when \ \frac{x}{2}+\frac{\pi}{3}=0 \therefore x= \frac{-2\pi}{3} \ then \ \boxed{Max(-\frac{2\pi}{3},7)} \\ \\ Minimum \ point:when \ \frac{x}{2}+ \frac{\pi}{3}= -\pi \therefore x=-\frac{8\pi}{3} \\ or \ x=-\frac{8\pi}{3}+4\pi=\frac{4\pi}{3} \then \ \boxed{min(\frac{4\pi}{3},-3)}$

9. Example for the tangent function.

As shown in Figure 3 we have the graph of the following function:

$f_{2}(x)=5tan(\frac{\pi x}{2}+\frac{\pi}{3})+2$

So the key features are:

$Amplitude: DNE \\ \\ Period:T= \frac{\pi}{\frac{1}{2}}=\boxed{2\pi} \\ \\ Phase \ shift:\boxed{\frac{\phi}{\omega}=\frac{2\pi}{3}} \\ \\ Vertical \ shift:\boxed{2} \\ \\ Maximum \ point: DNE \\ \\ Minimum \ point:DNE$

Describe how to transform a trigonometric function from the standard trig function f(x) = sin x, f(x

Describe how to transform a trigonometric function from the standard trig function f(x) = sin x, f(x

Answer from: Quest

$answer \:$ paul's with a y-intercept of $ 2856

Answer from: Quest

starting with:

5 + x - 12 = 2x - 7

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)

5 - 12 = x - 7 (combine and cancel x's from both sides)

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)

0 = x (add numbers together on both sides)

so in order for the equation to be true, x must be 0.

trying x = -0.5

5 + x - 12 = 2x - 7 (original equation)

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)

5 + -0.5 - 12 = -1 - 7 (did multiplication)

-7.5 = -8 (did addition on both sides)

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.

trying x = 0

5 + x - 12 = 2x - 7 (original equation)

5 + 0 - 12 = 2(0) - 7 (plug in value of x)

5 + 0 - 12 = 0 - 7 (did multiplication)

-7 = -7 (did addition on both sides)

since -7 equals -7, that demonstrates that the value of 0 for x is correct.

trying x = 1

5 + x - 12 = 2x - 7 (original equation)

5 + 1 - 12 = 2(1) - 7 (plug in value of x)

5 + 1 - 12 = 2 - 7 (did multiplication)

-6 = -5 (did addition on both sides)

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

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