write a cosine function that has a midline of $y = 4$, an amplitude of $5$ and a period of $\frac{7pi}{3}$…

Answer:

$f(x)=5\cos\left(\frac{6}{7}x\right)+4$

Explanation:

Step1: Recall the general form of cosine function

The general form of a cosine function is $y = A\cos(Bx - C)+D$.

Step2: Determine the values of $A$, $B$, $C$, and $D$

• The amplitude is $|A|$. Given amplitude $= 5$, so $A = 5$.
• The mid - line is $y=D$. Given mid - line $y = 4$, so $D = 4$.
• The period formula is $T=\frac{2\pi}{|B|}$. Given $T=\frac{7\pi}{3}$, then $\frac{7\pi}{3}=\frac{2\pi}{|B|}$. Solving for $B$: $$ \begin{align*} |B|&=\frac{2\pi}{\frac{7\pi}{3}}\ |B|&=\frac{6}{7} \end{align*} $$ We can take $B=\frac{6}{7}$ (since there is no phase shift information, we assume $C = 0$).

Step3: Write the function

Substitute $A = 5$, $B=\frac{6}{7}$, $C = 0$, and $D = 4$ into the general form $y = A\cos(Bx - C)+D$. We get $f(x)=5\cos\left(\frac{6}{7}x\right)+4$.