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

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

Answer from: Quest

if you need just tell me and i will come to the rescue.

step-by-step explanation: