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𝑓𝑡
1. The average speed for the team is about 93𝑚𝑝ℎ, which is 136.4 𝑠 . My height is 6𝑓𝑡.
The equation is ℎ(𝑡) = −16𝑡 2 + 136.4𝑡 + 6.
𝑏
A function 𝑎𝑡 2 + 𝑏𝑡 + 𝑐, 𝑎 < 0, has a minimum at 𝑡0 = − 2𝑎.
For our function it is 𝑡0 =
136.4
32
≈ 4.3 (𝑠) and ℎ(𝑡0 ) = −16𝑡02 + 136.4𝑡0 + 6 ≈ 296.7 (𝑓𝑡),
so the vertex is approximately (𝟒. 𝟑, 𝟐𝟗𝟔. 𝟕).
The roots may be found using quadratic formula:
−136.4 ± √(136.4)2 + 4 ∙ 16 ∙ 6
𝑡1,2 =
≈ −0.04 and 8.6 (𝑠).
−32
Only the positive root has sense in this situation, 𝟖. 𝟔 𝒔.
2. Problem 6.
The parent function is 𝑦 = |𝑥|. Its graph consists of two rays having the same angle with
the x-axis. To get the function needed we'll multiply this parent function by some negative
constant and move it (its vertex) into the point of reflection (𝑥𝑟 , 𝑦𝑟 ).
The point of reflection has the y-coordinate of 8 (ft). The angle of incidence is equal to the
𝑥 −2.5
angle of reflection; therefore, their tangents are also equal. The tangents are 08−2 and
9.5−𝑥0
8−2
, so 𝑥0 =
2.5+9.5
2
= 6.
This shows that the function has the form 𝑓(𝑥) = 𝑎|𝑥 − 6| + 8
for some (negative) constant 𝑎.
To find 𝑎, substitute the coordinates of the hole: 2 = 𝑎|9.5 − 6| + 8 = 3.5𝑎 + 8,
6
12
𝑎 = − 3.5 = − 7 . Thus the equation for the path of the ball is
𝒇(𝒙) = −
𝟏𝟐
|𝒙 − 𝟔| + 𝟖, 2.5 ≤ 𝑥 ≤ 9.5.
𝟕
(We may check whether this graph starts at the starting point.
12
12
Indeed, 𝑓(2.5) = − 7 |2.5 − 6| + 8 = − 7 ∙ 3.5 + 8 = −6 + 8 = 2).
3. Problem 10.
a. The rowed distance is √22 + 𝑥 2 , so the time on water is
1
2
√22 + 𝑥 2 .
The walked distance is √12 + (3 − 𝑥)2 , so the time on land is
𝟏
𝟏
1
4
√12 + (3 − 𝑥)2 .
This way the total time is 𝑻 = 𝟐 √𝟐𝟐 + 𝒙𝟐 + 𝟒 √𝟏𝟐 + (𝟑 − 𝒙)𝟐 .
b. x may be any real number. But it is sufficient to consider 𝒙 ∈ [𝟎, 𝟑]
because for 𝑥 < 0 or 𝑥 > 3 the time would not be minimal.
c. The graph (desmos.co...
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