Chapter 4 Review
1a) −5𝑥 + 4 = 0
-4
-4
- 5x =- 4
−𝟒
𝟒
Answer: x = −𝟓 or 𝟓
1b) −5𝑥 + 4 > 0
-4 -4
- 5x > -4 (divide both sides by – 5 and
switch
direction of sign because divide by
Negative
𝟒
Answer: 𝒙 < 𝟓
1c) 3x – 12 = - 5x + 4
+5x + 12 + 5x + 12
8x = 16
Answer: x = 2
1d) -5x + 4 < 3x – 12
-3x -4 -3x - 4
-8x < -16 (divide by -2 and switch direction
of sign since divide by negative
Answer: x > 2
2a) create two points. My points will be of the form (Time, Value)
(8,2000) and (0,50000) are my points.
The second point says when the tv’s are 0 years old they are worth $50,000.
now find m
𝑚=
𝑚=
50000−2000
0−8
48000
= −6000
−8
so m = -6000
Use m = -6000, x1 = 8, y1 = 2000
y – 2000 = -6000(x – 8)
y – 2000 = -6000x + 48000
+2000
+2000
Acceptable answer: y = -6000x + 50,000
answer may be written using function notation. I can also change the letter x to the letter t to stand for
time.
better answer #2a: V(t) = -6000t + 50,000
2b) replace the t with 6 and find the value
V(6) = -6000*6 + 50000
V(6) = 14000
Answer #2b: $14,000
3a) Create two points of the form (Price, units rented)
(1000 , 1000) (1300, 900)
Now find m:
𝑚=
900−1000
1300−1000
Let m =
−1
3
−100
300
=
=
−1
3
, x1 = 1000, y1 = 1000
y – 1000 =
−1
(x-1000)
3
y – 1000 =
−1
𝑥
3
+1000
+
1000
3
+
Answer #3a: 𝒚 =
3000
3
−𝟏
𝟒𝟎𝟎𝟎
𝒑+ 𝟑
𝟑
(I used p for price instead of x)
3b) plug $1200 for p and solve for y
𝑦=
−1
3
∗ 1200 +
4000
3
I got 933.333333 when I used my calculator. I need to round this answer.
Answer #3b: 933 units will be rented
3c) Let y = 800 and solve for p.
800 =
−1
𝑝
3
+
4000
3
−3 ∗ 800 = −3 ∗
−1
𝑝
3
+ −3 ∗
4000
3
-2400 = 1p – 4000
1600 = p
Answer #3c: a price of $1600 will cause 800 units to be rented
4) Problem 4 has been deleted
5a) 2f(x+1) + 3 = 2(x+1)2 + 3
the +1 inside the parenthesis shifts 1 unit to the left,
the + 3 after the parenthesis moves up 3 units
the 2 in front of the parenthesis stretches the graph or you may say it makes it narrower
Answer #5a: 2f(x+1) + 3 = 2(x+1)2 + 3
left 1, up 3, stretched or narrower
5b)
x
1
0
-1
-2
-3
h(x)
11
5
3
5
11
5c) domain for all parabolas (−∞, ∞), to compute the range the graph needs to be extended to the top
of the y-axis. Range will be of the form [y-coordinate of bottom point, ∞)
Answer #5c: Domain (−∞, ∞) Range [𝟑, ∞)
5d) The graph is increasing to the right of the vertex, and decreasing to the left of the vertex:
Answer #5d: increasing (−𝟏, ∞)
decreasing (−∞, −𝟏)
5d) there is no maximum point
5e) The vertex is the minimum point
Answer #5e: a local minimum occurs at x = -1, the minimum value is y = 3
1
1
6a) 2 𝑓(𝑥 + 2) − 4 =2 (𝑥 + 2)2 − 4
The + 2 in the parenthesis shifts left 2, the ½ in front makes the graph wider
the – 4 moves down 4
the ½ compresses vertically (you are not responsible for this part of the answer)
Answer #6a:
𝟏
𝒇(𝒙 +
𝟐
𝟏
𝟐) − 𝟒 =𝟐 (𝒙 + 𝟐)𝟐 − 𝟒
left 2, down 4, compressed or wider
6b)
x
0
-1
-2
-3
-4
g(x)
-2
-3.5
-4
-3.5
-2
6c) domain for all parabolas(−∞, ∞), to compute the range the graph needs to be extended to the top
of the y-axis. Range will be of the form [y-coordinate of bottom point, ∞)
Answer #6c: Domain (−∞, ∞) Range [−𝟒, ∞)
6d) The graph is increasing to the right of the vertex, and decreasing to the left of the vertex:
Answer #6d: increasing (−𝟐, ∞)
decreasing (−∞, −𝟐)
6d) there is no maximum point
6f) The vertex is the minimum point
Answer #6e: a local minimum occurs at x = -2, the minimum value is y = -4
7a) -3f(x) + 8 = -3x2 + 8
The (-) in front of the 3 reflects over x-axis
the + 8 moves up 8
the 3 stretches or makes narrower
Answer #7a: -3f(x) + 8 = -3x2 + 8
reflected over x-axis, up 8 , stretched or narrower
7b)
x
2
1
0
-1
-2
m(x)
-4
5
8
5
-4
7c) domain for all parabolas (−∞, ∞), to compute the range the graph needs to be extended to the
bottom of the y-axis. Range will be of the form (−∞, 𝑦 − 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑡𝑜𝑝 𝑝𝑜𝑖𝑛𝑡]
Answer #7c: Domain (−∞, ∞) Range (−∞, 𝟖]
7d) The graph is increasing to the left of the vertex, and decreasing to the rigth of the vertex:
Answer #7d: decreasing (𝟎, ∞)
increasing (−∞, 𝟎)
7e) The vertex is a maximum point
Answer #7e: a local maximum occurs at x = 0, the maximum value is y = 8
7f) Answer #7f: there is no minimum point
8a) group x’s:
1
m(x) = (x2 - 4x ) + 5
2
Find C = (2 ∗ −4) = (−2)2 = 4
Add and subtract C
m(x) = (x2 – 4x + 4) + 5
–4
factor / combine
Answer #8a: m(x) = (x-2)2 + 1
9a)
Group x’s h(x) = (-4x2 + 16x ) - 6
Factor out -4 h(x) = -4(x2 – 4x ) -6
1
2
Find C = (2 ∗ −4) = 4
Add 4 inside parenthesis , this is really like
subtracting 16 if you clear the parenthesis
the +4 will become – 16, so I will add 16
outside the parenthesis
h(x) = -4(x2 – 4x +4) - 6 + 16
factor and combine
Answer #9a: h(x) = -4(x-2)2 + 10
9b)
10a) let h = and solve for t
0 = -16t2 + 80t + 6 (divide by (-2) to make numbers smaller)
0 = 8t2 – 40t – 3 (use quadratic formula with a = 8, b = -40, c = -3
𝑡=
−(−40)±√(−40)2 −4(8)(−3)
2∗8
𝑡=
40+41.18
16
𝑜𝑟
=
40±√1696
16
=
40±41.18
16
40−40.18
16
t = 5.07 or -.01(this can’t be the answer as time must be positive)
Answer #10a: 5.07 seconds
10b) ball reaches maximum height at x-coordinate of vertex
a = -16 b = 80
−𝑏
−80
−80
x-coordinate of vertex = 2𝑎 = 2∗−16 = −32 = 2.5
Answer #10b: ball reaches maximum height at 2.5 seconds
10c) maximum height occurs at y-coordinate of vertex
h = -16(2.5)2 +80(2.5)+6
h = 106
Answer #10c: Maximum height 106 feet
11a) replace h with 80 and solve for t
80 = -16t2 + 96t
-80
-80
0 = -16t2 + 96t - 80 (divide by -16)
0 = t2 – 6t + 5
0 = (t – 1) (t – 5)
Answer #11a: height is 80 feet at 1 and 5 seconds
11b) replace h with 0 and solve for t.
0 = -16t2 + 96t
0 = -16t( t – 6)
-16t = 0
t–6=0
t=0
t=6
t = 0 (ignore as this is when it started, not landed) or t = 6
Answer #11b: it will take 6 seconds for object to return to the ground
11c) maximum height occurs at y-coordinate of vertex
x-coordinate of vertex =
−𝑏
2𝑎
=
−96
2∗−16
=3
y-coordinate of vertex = -16(3)2 + 96(3) = 144
answer #11c: 144 feet is maximum height
12a) plot points (1,5) (2,20) (3, 40) (4,25) and (5,5) do not need to show this graph
12b) use linreg and quadreg on calculator to get these answers.
Answer #12b: linear equation y = .5x + 12.25
quadratic equation y = -5.83x2 + 35.5x – 25.67
12c) the quadratic equation is clearly the better fit
12d) this graph does not need to be drawn
12e) max height occurs at y-coordinate of vertex
x-coordinate of vertex =
−𝑏
2𝑎
=
−35.5
2∗−5.83
= 3.04
y-coordinate of vertex = -5.83(3.04)2 + 35.5(3.04) – 25.67 = 28.37
(you may also use the maximum feature on your calculator to get this answer.)
answer: maximum height is about 28.37 feet (according to the equation)