Question

Question 13 and 14 please! Answer and show work :)

Answer 1

Problem 13)
`f(x)=4\,sin((1)/(2) x+(2)/(3)\pi )-2`

Problem 14)
`f(x)=cotan(x+(1)/(3) \pi)+2`

Explanation:

Recall how transformations affect the graph of the sine function, and how such is conveyed into the parameters A, B, C, and D that could be included in the general form of the function:

`f(x)=A\,sin(Bx+C)+D`

where the Amplitude of the transformed sine function is the absolute value of the multiplicative parameter A:

Amplitude =
`|A|`

The period is (which for sin(x) is
`2\pi`) is modified by the parameter B in the following manner:

Period =
`(2\pi)/(B)`

Where the phase shift is introduced as:

Phase shift =
`-(C)/(B)`.

and finally any vertical shift is included by the constant D (positive means shift upwards in D many units, and negative means shift downwards D units)

Therefore, to have a sine function with the requested characteristics, we work on the value of the parameters A, B, C, and D one at a time:

1) Amplitude =
`|A|=4` then we use parameter A = 4

`f(x)=4\,sin(Bx+C)+D`

2) Period
`4\pi`, then we work on the parameter B:

Period =
`(2\pi)/(B)`

`4\pi=(2\pi)/(B)\\B*4\pi=2\pi\\B=(2\pi)/(4\pi) \\B=(1)/(2)` which transforms the function into:

`f(x)=4\,sin((1)/(2) x+C)+D`

3) phase-shift =
`-(4)/(3) \pi`

Then knowing that B=
`(1)/(2)`, we work on the value of parameter C:

Phase shift =
`-(C)/(B)`

`-(4)/(3) \pi=-(C)/(B) \\-(4)/(3) \pi=-(C)/( (1)/(2) )\\-(4)/(3)* (1)/(2) \pi=-C\\C=(2)/(3) \pi`

Therefore the function gets transformed into:

`f(x)=4\,sin((1)/(2) x+(2)/(3)\pi )+D`

4) and finally the vertical shift of negative two units, that gives us the value D = -2

The complete transformed function becomes:

`f(x)=4\,sin((1)/(2) x+(2)/(3)\pi )-2`

Now for problem 14, recall that the cotangent function is the reciprocal of the tangent function, therefore, their periodicity is the same:
`\pi`

since you are asked for a cotangent function of period
`\pi` as well, there is no multiplication parameter "B" needed (so we keep it unchanged - equal to one). B = 1

Then for the phase-shift which we want it to be
`-(1)/(3) \pi`, we set the condition:

`-(1)/(3) \pi=-(C)/(B) \\-(1)/(3) \pi=-(C)/(1)\\-(1)/(3) \pi=-C\\C=(1)/(3) \pi`

And insert such in the cotangent general form:

`f(x)=cotan(x+(1)/(3) \pi)+D`

and finally include the desired vertical shift of 2 units:

`f(x)=cotan(x+(1)/(3) \pi)+2`