Question

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

Answer 1

Answer:

Problem 13) $f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )-2$

Problem 14) $f(x)=cotan(x+\frac{1}{3} \pi)+2$

Step-by-step 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 = $\frac{2\pi}{B}$

Where the phase shift is introduced as:

Phase shift = $-\frac{C}{B}$.

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 = $\frac{2\pi}{B}$

$4\pi=\frac{2\pi}{B}\\B*4\pi=2\pi\\B=\frac{2\pi}{4\pi} \\B=\frac{1}{2}$ which transforms the function into:

$f(x)=4\,sin(\frac{1}{2} x+C)+D$

3) phase-shift = $-\frac{4}{3} \pi$

Then knowing that B=$\frac{1}{2}$, we work on the value of parameter C:

Phase shift = $-\frac{C}{B}$

$-\frac{4}{3} \pi=-\frac{C}{B} \\-\frac{4}{3} \pi=-\frac{C}{ \frac{1}{2} }\\-\frac{4}{3}* \frac{1}{2} \pi=-C\\C=\frac{2}{3} \pi$

Therefore the function gets transformed into:

$f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )+D$

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

The complete transformed function becomes:

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

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 $-\frac{1}{3} \pi$, we set the condition:

$-\frac{1}{3} \pi=-\frac{C}{B} \\-\frac{1}{3} \pi=-\frac{C}{1}\\-\frac{1}{3} \pi=-C\\C=\frac{1}{3} \pi$

And insert such in the cotangent general form:

$f(x)=cotan(x+\frac{1}{3} \pi)+D$

and finally include the desired vertical shift of 2 units:

$f(x)=cotan(x+\frac{1}{3} \pi)+2$