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**Unformatted text preview: **asing on 5 , ∞
6
Vertex 5 , 73 is a minimum
6 12
Axis of symmetry x = 5
6 73
12 y
6
5
4
3
2
1 −1
−1
−2
−3 1 2 3 x 152 Linear and Quadratic Functions
(h) f (x) = x2 − 1
100 x − 1 =
√
1+ 40001
200 x− 40001
12
200 √ − 40000
1− 40001
200 x-intercepts
and
y -intercept (0, −1)
Domain: (−∞, ∞)
40001
Range: − 40000 , ∞
1
Decreasing on −∞, 200
1
Increasing on 200 , ∞
40001
1
Vertex 200 , − 40000 is a minimum8
1
Axis of symmetry x = 200
2. y = |1 − x2 | 7
6
5
4
3
2
1
−2 −1 3. y y
8 1 2 x √
√
3 − 7 −1 + 7
,
,
2
2 √
√
3 + 7 −1 − 7
,
2
2 7
6
5
4
3
2
1
−2 −1 1 2 x 5. (a) The applied domain is [0, ∞).
(d) The height function is this case is s(t) = −4.9t2 + 15t. The vertex of this parabola
is approximately (1.53, 11.48) so the maximum height reached by the marble is 11.48
meters. It hits the ground again when t ≈ 3.06 seconds.
(e) The revised height function is s(t) = −4.9t2 + 15t + 25 which has zeros at t ≈ −1.20 and
t ≈ 4.26. We ignore the negative value and claim that the marble will hit the ground
after 4.26 seconds.
(f) Shooting down means the initial velocity is neg...

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