Answer :
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]
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]
