Answer: [tex]y=0.4 \cos \left(\frac{2\pi}{0.9} (x-0.8) \right)+1.1[/tex]
Step-by-step explanation:
The amplitude of the function is [tex]\frac{1.5-0.7}{2}=0.4[/tex].
The midline of the graph is [tex]y=\frac{1.5+0.7}{2}=1.1[/tex].
The period of the graph is [tex]1.7-0.8=0.9[/tex], making the frequency [tex]\frac{2\pi}{0.9}[/tex].
Therefore, the equation of the graph is of the form [tex]y=0.4 \cos \left(\frac{2\pi}{0.9} (x-c) \right)+1.1[/tex] for some constant [tex]c[/tex].
Since the graph passes through [tex](1.25, 0.7)[/tex], we have that:
[tex]0.7=0.4 \cos \left(\frac{2\pi}{0.9} (1.25-c) \right)+1.1\\\\0.4 \cos \left(\frac{2\pi}{0.9} (1.25-c) \right)=-0.4\\\\\cos \left(\frac{2\pi}{0.9} (1.25-c) \right)=-1\\\\\frac{2\pi}{0.9} (1.25-c)=\pi\\\\\frac{2}{0.9} (1.25-c)=1\\\\1.25-c=0.45\\\\c=0.8[/tex]
Therefore, the equation of the graph is [tex]y=0.4 \cos \left(\frac{2\pi}{0.9} (x-0.8) \right)+1.1[/tex].