1. The total annual enrollment (in millions) in U.S. elementary schools for the years 1975-1996 is given by the model

E = -0.058x^2 -1.162x +50.604----------(i)

Where x=0 corresponds to 1975, x=1 corresponds to 1976, and so on. For this period, when was ennrollment the lowest? To the nearest tenth of a million, what was the enrollment?

If equation (i) is correct, then that is a vertical parabola that opens downward---because of the negative coeefficient of the x^2. So its vertex is a maximum point.

Its vertex is at

x = -b/2a = -(-1.162)/[2(-0.058)] = -10.017

That means the vertex is at year 1975 -10 = 1965. Which means the enrollment was highest in 1965.

The parabola further means after 1965, the enrollment decreases every year "parabolically"---not linearly.

Therefore, even without solving, the lowest enrollment in the interval [1975,1996] is in 1996, where the enrollment then is:

1996 -1975 = 21 years

E = -0.058(21^2) -1.162(21) +50.604

E = 0.624

To the nearest tenth,

E = 0.6 millions

E = 600,000 students only

Is that realistic?

In 1996, there were only about 600,000 students enrolled inallUS elementary schools?

The posted model is correct?

[In 1975, E = -0.058(0^2) -1.162(0) +50.604 = 50.604 millions elementary students enrolled.]

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2. A parabolic arch has an equation of

x^2 +20y -400 = 0

where x is measured in feet. Find the maximum height of the arch.

So, rearranging,

20y = -x^2 +400

y = -(1/20)x^2 +20 -----------(ii)

The parabola is vertical, opening downward, so its vertex is a maximum.

The vertex is at

x = -b/(2a) = -0/[2(-1/20)] = 0

So, the y at x=0 is

y = -(1/20)(0^2) +20 = 20.

Therefore, the maximum height of the arch is 20 ft. -------------answer.