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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First, "graph" means graph of the given function f(x). It is not necessary to include the graph with the question because at this level, either the user can "see" the graph mentally, or use technology (calculator or software) to graph the function.
(a) The Leading coefficient test for polynomial functions
The test is as follows. The sign of the leading coefficient (LC, coefficient of the highest power of the variable) together with the degree (odd or even) of the polynomial function tells us the end behaviour of the function according to the following table:
LC/degree Even Odd
Positive +inf/+inf -inf/+inf
Negative -inf/-inf +inf/-inf
It can be easily memorized by comparing with a line or a parabola
y=x (positive leading coefficient, odd degree) goes from -inf to +inf
y=x^2 (positive leading coefficient, even degree) goes from +inf to +inf.
Any negative leading coefficient will just invert the signs.
For the given polynomial function
f(x) = x^2(x+2) = x^3+x^2
The leading coefficient is +1 (positive), and degree is three (odd)
So it behaves the same way as a straight line y=x, so the end behviours are
{-inf, +inf} as x-> {-inf, +inf} respectively.
(b) x-intercepts
Since f(x)=x^2(x+2), the x-intercepts are x=0 (multiplicity 2) and x=-2.
=>
From the left, graph is increasing and crosses x=-2 once from negative to positive, after which graph is decreasing. At x=0, because of the root with multiplicity 2, graph touches the origin (0,0) and turns back towards +inf.
(c) the y-intercept
from the polynomial f(x)=x^2(x+2)=x^3+x^2 = x^3+x^2+0,
we see that the y-intercept =0 because the "initial value" or the constant term is zero.
(a) The Leading coefficient test for polynomial functions
The test is as follows. The sign of the leading coefficient (LC, coefficient of the highest power of the variable) together with the degree (odd or even) of the polynomial function tells us the end behaviour of the function according to the following table:
LC/degree Even Odd
Positive +inf/+inf -inf/+inf
Negative -inf/-inf +inf/-inf
It can be easily memorized by comparing with a line or a parabola
y=x (positive leading coefficient, odd degree) goes from -inf to +inf
y=x^2 (positive leading coefficient, even degree) goes from +inf to +inf.
Any negative leading coefficient will just invert the signs.
For the given polynomial function
f(x) = x^2(x+2) = x^3+x^2
The leading coefficient is +1 (positive), and degree is three (odd)
So it behaves the same way as a straight line y=x, so the end behviours are
{-inf, +inf} as x-> {-inf, +inf} respectively.
(b) x-intercepts
Since f(x)=x^2(x+2), the x-intercepts are x=0 (multiplicity 2) and x=-2.
=>
From the left, graph is increasing and crosses x=-2 once from negative to positive, after which graph is decreasing. At x=0, because of the root with multiplicity 2, graph touches the origin (0,0) and turns back towards +inf.
(c) the y-intercept
from the polynomial f(x)=x^2(x+2)=x^3+x^2 = x^3+x^2+0,
we see that the y-intercept =0 because the "initial value" or the constant term is zero.