1) the curve of has vert asy: x = 20 and horiz asy: y = 1000

We need to find 2d where d is the distance between C, the center of f(x)
and the intersection point of f(x) with the line, slope = 1, through C.
Equation of the line is (y - 1000) = (x - 20) t y = x + 980.
Substituting this for f(x), gives

(x - 20)² = 1600 t x = -20 or x = 60
Substituting x = 60 we get y = 40
Now, d is the distance between (20, 1000) and (60, 40) = 960.8 meters
At the closest point, the jets come within 1921.7 meters of each other
(but it looks closer in the rear view mirror!).

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2) a) Since the zeros are -2 and 6, h = 2 (midpoint), so (h, k) = (2, 4).
a = slope of VB = -1 so f(x) = - | x - 2 | + 4
b)

c) - | x - 2 | + 4 ¥ 2 t - | x - 2 | ¥ - 2 -- divide both sides by -1 (reverse inequality)
| x - 2 | [ 2 t - 2 [ x - 2 [ 2 t 0 [ x [ 4

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3) a) Slope of AB = a = -8 / 15 so f(x) = -8/15 | x - 7.5 | + 4.

b)

domain: [0, 15]	range: [0, 4]	increasing: [0, 7.5]	decreasing: [7.5, 15]
f(x) > over domain	zeros: 0 and 15	maximum: 4

c)

We compare similar triangles AEO and ACG.
We know that so, 4 - x = 2.4 meters.

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4) a)

b) The inverse function is:

c) The plane's altitude is < 1000 meters for 15.16 seconds. (solve h(t) < 1000).

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5)

vertical asymptote	horizontal asymptote	y-intercept	zero
a) x = 1	y = 3	(0, 1)	(1/3, 0)
b) x = 3.5	y = 2	(0, 5/7)	(5/4, 0)
c) x = ½	y = - ¾	(0, 1)	(-2/3, 0)
d) x = 1	y = 1	(0, - 1)	(- 1, 0)

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6) Set x = 32 in each function, multiply by 3 for 3 years.

p1 = $300.38

p2 = $299.29

p3 = $286.50

p4 = $295.80

The best deal is company C.

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7)

a)	b) 75x + 100 < 85x t x > 10 hours
c) 80x = 75x + 100 t x = 20 hours	d) no, it's always $75 plus something.

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8)

a) g [f(x) = g(x + 1) = x² - 4	b) h [f(x)] = h(x + 1) = (2x + 1)½ .
c) k [f(x)] = k(x + 1) = x / (2x + 3)	d) it's a quadratic function

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