p72 #1, 3, 5, 8, 9, 11, 12, 15, 19, 21, 23 Quadratic Chapter
Review
1a) The sequence forms an arithmetic sequence because there is a common difference
b) t1 = 4, d = 7
tn = t1 + d(n-1)
tn = 4 + 7(n-1) = 4 + 7n – 7 = 7n – 3
t50 = 7(50) – 3 = 350 – 3 = 347
c) 1000 = 7n – 3
1000 + 3 = 7n
1003/7 = n
143.2 = n
t144 is slightly more than n.
3. There are many possible solutions for this question. One solution would be {3, 7, 15, 27, 43 }.
5. 100 = 3(x + 3x) + 3(y + 2y)
100 = 12 x + 9 y
9y = 100 – 12 x
y = 100/9 – 12/9 x
A= l x w
A = (x + 3x) (y + 2y)
A = (4x)(3y)
A = (4x) 3(100/9 – 12/9 x)
A = (4x)(100/3 – 4x)
A = 400x/3 – 16x2
Use a graphing calculator to find the rest
8.
a)
b)
c)
d)
9.
a) Vertical Stretch: 3
Vertex: (-2, 9)
Axis of Symmetry: x = -2
Minimum Value: 9
b) Vertical Stretch: 1/2
Vertex: (1, -1/2)
Axis of Symmetry: x = 1
Minimum Value: -1/2
c) Vertical Stretch: -4
Vertex: (-1, 5)
Axis of Symmetry: x = -1
Maximum Value: 5
d) Vertical Stretch: -1
Vertex: (-1, 5)
Axis of Symmetry: x = -1
Maximum Value: 5
11. 4x + y = 120
y = 120 - 4x
Product = (x)(120 - 4x)
Product = 120x – 4x2
To find the max product we need the vertex.
So the x-coordinate of the vertex tells us the x-value that makes the maximum product.
We still need the y value that will make the maximum product.
y = 120 - 4x
y = 120 – 4(15)
y = 120 – 60 = 60
So x = 15 and y = 60. The maximum product is 900 (the k value).
12. You know the x-coordinate of the vertex (h) and two points on the parabola (two x and y values).
Create two equations with two unknows (k and a) and solve for k (the y-coordinate of the vertex).
y = 18
15.
a)
b)
c)
d)
e)
f)
g)
19.
a) 2 x-intercepts; 0.67 and 2.25
b) no x-intercepts
c) two x-intercepts; -0.69 and 8.69
d) two x-intercepts; -0.33 and 0.5
e) one x-intercept; -0.25
21. At x = 20, the rocket has an altitude 90 m. It clears the building by 5 m.
23. Let x = length of the court
Let y = width of the court
The court is 18.21 m wide by 36.79 m long.