Answer:
Step-by-step explanation:
Rewrite the quadratic using the method of completing the square:
From this form it is immediately clear that the vertex of the parabola is at x=1 and its y-intercept at -6.
Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates
1 answer:
Answer:
W(t) = 0.9cos(2πt/366) + 8.2
Step-by-step explanation:
W(t) = a cos(bt) + d
1. Calculate the phase shift, b
At t= 0, the dog is at maximum weight, so the cosine function is also at a maximum.
The cosine function is not shifted, so b = 1.
W(t) = a cos t + d
2. Calculate d
The dog's average weight is 8.2 kg, so the mid-line d = 8.2.
W(t) = a cos t + 8.2
3. Calculate a
The dog's maximum weight is 9.1 kg.
The deviation from the average (the amplitude, a) is 9.1 kg - 8.2 kg = 0.9 kg.
W(t) = 0.9cos t + 8.2
3. Calculate t
The period p = 2π/b = 2π/1 = 2π
From t = 0 to t = 91.25 da is one-quarter of a period, so
p = 4 × 91.25 da = 365 da = 2π rad
The conversion factor is 1 da =2π/365 rad
The function with t in radians is
W(t) = 0.9cos(2πt/365) + 8.2
The figure below shows the graph of the function.
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