write a cosine function that has an amplitude of 2, a midline of ( y = 4 ) and a period of ( \frac{6}{7}…

Answer:

( f(x)=2\cos\left(\frac{7\pi}{3}x\right)+4 )

Explanation:

Step1: Recall the general form of cosine function

The general form of a cosine function is ( y = A\cos(Bx - C)+D ), where ( A ) is the amplitude, ( \frac{2\pi}{|B|} ) is the period, ( C ) is the phase - shift (we assume ( C = 0 ) as there is no information about phase - shift), and ( D ) is the mid - line.

Step2: Determine the value of ( A )

Given that the amplitude ( |A|=2 ). We can take ( A = 2 ) (the sign of ( A ) affects the reflection, and without information about reflection, we can take the positive value).

Step3: Determine the value of ( D )

Given that the mid - line ( y = D), and ( y = 4), so ( D = 4 ).

Step4: Determine the value of ( B )

We know that the period ( T=\frac{2\pi}{|B|}). Given ( T=\frac{6}{7}). Then (\frac{6}{7}=\frac{2\pi}{|B|}). Solving for ( B ): [ \begin{align*} |B|&=\frac{2\pi\times7}{6}\ |B|&=\frac{7\pi}{3} \end{align*} ] We can take ( B=\frac{7\pi}{3}) (since the period formula only depends on the magnitude of ( B ) for non - phase - shifted functions and we can assume ( B>0 ) without loss of generality).

Substitute ( A = 2), ( B=\frac{7\pi}{3}), ( C = 0), and ( D = 4) into the general form ( y = A\cos(Bx - C)+D), we get ( f(x)=2\cos\left(\frac{7\pi}{3}x\right)+4 ).