The function M(x)=-0.287x^3+8.8^2-59.843x+220.7 describes the incidence of measels (per 100,000) for the period 1940-1960 (x=0 for 1940). In what year was the greatest incidence of measels reported? According to the definition of M (x), what is the y intercept?Identify periods of increasing/decreasing frequency of the disease. If the function continues to model the disease beyond 1960, when did the incidence of measel approximate zero? What are some variables that may have affected the incidence of measels over the period 1940-60?
I assume you have had calculus.
M'=0=-0.861x2+17.6x-59.843
Then, use the quadratic formula to solve for x to find the greatest year.
I will be happy to critique your thinking on the other questions.
To find the year when the greatest incidence of measles was reported, we need to find the maximum value of the function M(x). We can achieve this by finding the critical points of the function and evaluating M(x) at these points and at the endpoints of the given period.
To find the critical points, we take the derivative of the function M(x) with respect to x, which gives us:
M'(x) = -0.861x^2 - 59.843.
Setting M'(x) equal to zero and solving for x, we get:
-0.861x^2 - 59.843 = 0.
This is a quadratic equation, which can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a,
where a = -0.861, b = 0, and c = -59.843.
Calculating the discriminant, b^2 - 4ac, we get:
(-59.843)^2 - 4(-0.861)(0) = 3584.069.
Since the discriminant is positive, there are two real solutions to the quadratic equation. Solving for x, we get:
x = (√(3584.069)) / -1.722 (approximately -34.392),
or
x = -(-√(3584.069)) / 1.722 (approximately 34.392).
The period given is from x = 0 for 1940, so only the positive value of x (x = 34.392) is feasible in this case.
To find the corresponding year, we add 1940 to the value of x, so the year when the greatest incidence of measles was reported is approximately 1974.
Now, to determine the y-intercept, we need to find the value of M(x) when x = 0 (which represents the year 1940). Substituting x = 0 into the function M(x), we get:
M(0) = -0.287(0)^3 + 8.8(0)^2 - 59.843(0) + 220.7 = 220.7.
Therefore, the y-intercept of the function M(x) is 220.7.
To identify periods of increasing/decreasing frequency of the disease, we can examine the sign of the derivative M'(x). Recall that we already found the derivative M'(x) = -0.861x^2 - 59.843.
If the derivative is positive (M'(x) > 0), then the function is increasing. If the derivative is negative (M'(x) < 0), then the function is decreasing.
Analyzing the sign of M'(x), we can see that the derivative is negative for values of x between the two critical points we found earlier (x = -34.392 and x = 34.392). Therefore, the disease frequency was decreasing during this period. Before and after these intervals, the derivative is positive, indicating an increasing disease frequency.
To approximate when the incidence of measles approached zero, we need to solve M(x) = 0. However, M(x) is a cubic function, and solving cubic equations may not yield nice exact solutions. In this case, we can use numerical methods or approximation techniques to estimate the value of x when M(x) is close to zero.
Regarding the variables that may have affected the incidence of measles over the period 1940-1960, we need additional information. Ideally, we would examine historical records and data related to measles outbreaks during this period to identify potential variables. Possible variables that may have influenced the incidence of measles could include vaccination rates, population density, healthcare access, public health policies, hygiene practices, and so on. However, without more specific information, we cannot identify which variables specifically influenced the incidence of measles during this time.