write a cosine function that has a midline of $y = 4$, an amplitude of 5 and a period of $\frac{pi}{6}$…

Answer:

$f(x)=5\cos(12x)+4$

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 parameters

• Amplitude: $|A|$. Given amplitude $= 5$, so $A = 5$.
• Midline: $y=D$. Given midline $y = 4$, so $D = 4$.
• Period formula: $T=\frac{2\pi}{|B|}$. Given $T=\frac{\pi}{6}$, then $\frac{\pi}{6}=\frac{2\pi}{|B|}$. Solving for $B$: $$ \begin{align*} \frac{\pi}{6}&=\frac{2\pi}{|B|}\ |B|\times\pi&= 12\pi\ |B|&=12 \end{align*} $$ We can take $B = 12$ (since we are not given a phase - shift, $C = 0$).

Step3: Write the function

Substitute $A = 5$, $B = 12$, $C = 0$, $D = 4$ into $y = A\cos(Bx - C)+D$. We get $y=5\cos(12x)+4$.