Y=-7/8xproportional or not proportionalconstant of proportionality: k=
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First, let's use the given information to determine the function's amplitude, midline, and period.
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
= Solution
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.