Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

\begin{tabular}{|c|c|c|c|}
\hline
Function & Amplitude & Midline [tex]$y=$[/tex] & Period \\
\hline
[tex]$3 \sin \left(\frac{3}{2} x-3\right)+5$[/tex] & 3 & 5 & [tex]$\frac{2 \pi}{3}$[/tex] \\
\hline
[tex]$\frac{2}{3} \cos \left(5 x+\frac{1}{2}\right)+6$[/tex] & [tex]$\frac{2}{3}$[/tex] & 6 & [tex]$\frac{2 \pi}{5}$[/tex] \\
\hline
[tex]$\frac{5}{3} \sin (7 x-3)-4$[/tex] & [tex]$\frac{5}{3}$[/tex] & -4 & [tex]$\frac{2 \pi}{7}$[/tex] \\
\hline
[tex]$8 \cos \left(\frac{4}{5} x+2\right)-3$[/tex] & 8 & -3 & [tex]$\frac{5 \pi}{2}$[/tex] \\
\hline
\end{tabular}


Sagot :

Let's solve this question step-by-step for each function.

1. Function: [tex]\(\frac{2}{3} \cos \left(5 x+\frac{1}{2}\right)+6\)[/tex]

- Amplitude:
The coefficient of [tex]\(\cos\)[/tex] gives the amplitude. Thus, the amplitude is [tex]\(\frac{2}{3}\)[/tex].

- Midline:
The constant term outside the cosine function determines the midline. Here, the midline is [tex]\(y = 6\)[/tex].

- Period:
The period of the cosine function [tex]\(\cos(kx)\)[/tex] is given by [tex]\(\frac{2\pi}{k}\)[/tex]. In this case, [tex]\(k = 5\)[/tex], so the period is [tex]\(\frac{2\pi}{5} \approx 1.2566370614359172\)[/tex].

2. Function: [tex]\(8 \cos \left(\frac{4}{5} x+2\right)-3\)[/tex]

- Amplitude:
The coefficient of [tex]\(\cos\)[/tex] gives the amplitude. Thus, the amplitude is [tex]\(8\)[/tex].

- Midline:
The constant term outside the cosine function determines the midline. Here, the midline is [tex]\(y = -3\)[/tex].

- Period:
The period of the cosine function [tex]\(\cos(kx)\)[/tex] is given by [tex]\(\frac{2\pi}{k}\)[/tex]. In this case, [tex]\(k = \frac{4}{5}\)[/tex], so the period is [tex]\(\frac{2\pi}{\frac{4}{5}} = \frac{2\pi \cdot 5}{4} = \frac{10\pi}{4} = \frac{5\pi}{2} \approx 7.853981633974483\)[/tex].

Thus, the completed table is:

[tex]\[ \begin{tabular}{|c|c|c|c|} \hline Function & Amplitude & \begin{tabular}{c} Midline \\ $y=$ \end{tabular} & Period \\ \hline $3 \sin \left(\frac{3}{2} x-3\right)+5$ & 3 & 5 & $\frac{2 \pi}{5}$ \\ \hline$\frac{2}{3} \cos \left(5 x+\frac{1}{2}\right)+6$ & $\frac{2}{3}$ & 6 & $1.2566370614359172$ \\ \hline$\frac{5}{3} \sin (7 x-3)-4$ & $\frac{5}{3}$ & -4 & \\ \hline $8 \cos \left(\frac{4}{5} x+2\right)-3$ & 8 & -3 & $7.853981633974483$ \\ \hline \end{tabular} \][/tex]