Linear And Quadratic Functions Worksheet Page 9
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Linear and Quadratic Functions
25. (a) The applied domain is [0,
).
2
(d) The height function is this case is s(t) =
4.9t
+ 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.
2
(e) The revised height function is s(t) =
4.9t
+ 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 negative so the height functions becomes
2
s(t) =
4.9t
15t + 25.
26. Make the vertex of the parabola (0, 10) so that the point on the top of the left-hand tower
where the cable connects is ( 200, 100) and the point on the top of the right-hand tower is
9
2
(200, 100). Then the parabola is given by p(x) =
x
+ 10. Standing 50 feet to the right of
4000
the left-hand tower means you’re standing at x =
150 and p( 150) = 60.625. So the cable
is 60.625 feet above the bridge deck there.
2
27. y = 1
x
3
7
1 +
7
3 +
7
1
7
28.
,
,
,
y
2
2
2
2
7
6
5
4
3
2
1
x
2
1
1
2
2
2
2
2
29. D(x) = x
+(2x+1)
= 5x
+4x+1, D is minimized when x =
, so the point on y = 2x+1
5
2
1
closest to (0, 0) is
,
5
5
2
m
m
+ 4
31. x =
y 10
32. x =
(y
2)
33. x =
2
2
3
16x + 9
v
v
+ 4gs
0
0
0
34. y =
35. y = 2
x
36. t =
2
2g
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