Chapter 11 Summary and Review: Discussion and Writing Exercises 329
35. a) We look for a function of the form N(x)=ax2+
bx +cwhere N(x) represents the number of live
births per 1000 women and xrepresents the age of
36. x2−10x+25=0
37. 2x2−6x+5=0
x=6±√36 −40
4=6±√−4
4=6±2i
4
38. Familiarize. Let band hrepresent the base and height,
respectively, in centimeters.
2bh. Substituting −h+ 38, we get a
h=−b
2a=−19
−1=19
19 cm.
State. The maximum area of 180.5 cm2occurs when the
base is 19 cm and the height is 19 cm.
39. Familiarize. Let xrepresent one of the numbers. Then
u=−19 or u =18
then √x=√324 = 18. Since 324+18
2=342
2= 171, the
Chapter 11 Discussion and Writing Exercises
(0,c).
2. If the leading coefficient is positive, the graph of the func-
2is 3 less than (or 3 units
4. Find a quadratic function f(x) whose graph lies entirely
5. No; if the vertex is off the x-axis, then due to symmetry
330 Chapter 11: Quadratic Equations and Functions
Chapter 11 Test
1. a) 3x2−4=0
2. x2+x+1=0
a=1,b=1,c=1
x=−1±√−3
2=−1±i√3
2=−1
2±i√3
2
3. x−8√x+7=0
4. 4x(x−2) −3x(x+1) =−18
x−2=0 or x −9=0
5. 4x4−17x2+15=0
Let u=x2and think of x4as (x2)2,oru2.
4u−5=0 or u −3=0
x=−4±2√6
2=2(−2±√6)
2·1=−2±√6
7. 1
4−x+1
2+x=3
4LCM is (4−x)(2 + x)
2+x+4−x=3
4(8+2x−x2)
8. x2−4x+1=0
Chapter 11 Test 331
9. s(t)=16t2
10. Familiarize. Let r= the speed of the boat in still water
and let t= the time of the trip upriver. Then 4 −t=
the time of the return trip downriver. We organize the
have two equations:
t=3
4−3
4r2−16 −3r−6=3r−6
r=3±√73
downriver is about 2.89 + 2, or 4.89 mph, and the time it
width, in cm. Then the perimeter of the board is 2l+2w
and the area is l·w.
quadratic function.
A=l·w= (14 −w)w=14w−w2,or−w2+14w
Carry out. We complete the square in order to find the
The vertex is (7,49). The coefficient of w2is negative, so
the graph of the function is a parabola that opens down.
Check. We could find the value of the function for some
12. V(T)=48T2
13. x2+5x+17=0
14. The solutions are √3 and 3√3.
332 Chapter 11: Quadratic Equations and Functions
15. V=48T2
Thus, we also have T=√3V
12 .
16. f(x)=−x2−2x
=−(x2+2x+1−1) 2
22
=1
2=1;
add 1 −1
17. f(x)=4x2−24x+41
add 9 −9
=4(x2−6x+9)+4(−9)+41
18. f(x)=−x2+4x−1
The y-intercept is (0,f(0)). Since
f(0) = −02+4·0−1=−1, the y-intercept is (0,−1).
19. Familiarize. Let xand yrepresent the numbers.
Translate. The difference of the numbers is 8, so we have
smaller than these function values. We could also use the
20. We look for a function of the form f(x)=ax2+bx +c.
Substituting the data points, we get
Chapter 11 Test 333
21. a) Substituting the data points, we get
18.5=a·02+b·0+c,
b) In 2011, x= 2011 −2000 = 11.
22. x2<6x+7
x2−6x−7<0
−17
We try test numbers in each interval.
A: Test −2,y=(−2 + 1)(−2−7) = 9 >0
23. x−5
A: Test −4,x−5
−5
3TRUE
x+3 <0
6−5
24. x−2
(x+ 3)(x−1) ≥0
Solve the related equation.
Find the numbers for which the rational expression is un-
defined.
A
B
C
D
B: Test 0, x−2
(3 + 3)(3 −1) ?0
1
25. x=i
2or x =−i
2
26. We look for a function f(x)=ax2+bx +c. We substitute
27. kx2+3x−k=0
First we substitute −2 for xand find k.
Cumulative Review Chapters 1 – 11 335
Cumulative Review Chapters 1 – 11
1. Let d= the distance that a hole in one shot would travel,
in yards. We use the Pythagorean theorem.
2. (4+8x2−5x)−(−2x2+3x−2) =
4. a2−16
5a−15 ·2a−6
a+4 =(a2−16)(2a−6)
(5a−15)(a+4)
5. y
y2−y−42 ÷y2
y−7=y
y2−y−42 ·y−7
y2
6. 2
m+1+3
m−5−m2−1
m2−4m−5
=2
m+1+3
m−5−m−1
m−5,LCD is (m+ 1)(m−5)
=−(m2−5m+6)
(m+ 1)(m−5)
7. We will use synthetic division.
(9x3+5x2+2)÷(x+2) = (9x3+5x2+0x+2)÷[x−(−2)]
8.
1
x−1
y
1
x−1
y
10. √9x2−36x+36 = 9(x2−4x+4) = 9(x−2)2=
336 Chapter 11: Quadratic Equations and Functions
12. 2√3−4√2
√2−3√6=2√3−4√2
√2−3√6·√2+3
√6
√2+3
√6
=2√6+6
√9·2−8−12√4·3
2−54
=√6+9
−26
−18i−6
9=−2i−2
3,or −2
3−2i
16. Using trial and error or the ac-method, we have
12x−15+6=3−x−1
12x−9=−x+2
25. F=mv2
r
F
26. 5−3(2x+1)≤8x−3
27. 3x−2<−6or x +3>9
28. |4x−1|≤14
−14 ≤4x−1≤14
29. 5x+10y=−10,(1)
30. 2x+y−z=9,(1)
Cumulative Review Chapters 1 – 11 337
Now multiply Equation (1) by 2 and add it to Equation (3)
x=1
Finally, substitute 1
2for xand 3 for yin Equation (2) and
31. 10x2+28x−6=0
3=n
33. 1
34. A=mh
m+a
35. √2x−1=6
36. √x−2+1=√2x−6
√2x−6)2
37. 16(t−1) = t(t+8)
338 Chapter 11: Quadratic Equations and Functions
38. x2−3x+16=0
39. 18
x+1−12
x=1
3,LCM is 3x(x+1)
40. P=√a2−b2
41. (x+ 3)(x+2)
(x−1)(x+1) <0
A: Test −4, (x+ 3)(x+2)
B: Test −2.5, (x+ 3)(x+2)
(x−1)(x+1) <0
−2.5(−0.5)
(x−1)(x+1) <0
(0+3)(0+2)
y
y
y
x
42. 4x2−25 >0
(2x+ 5)(2x−5) >0
43. Graph x+y=2.
We find some ordered pairs that are solutions of the equa-
44. Graph y≥6x−5.
First graph the equation y=6x−5 using a solid line since
45. Graph x<−3.
y<−3
0?−3 FALSE
46. Graph 3x−y>6,
47. Graph f(x)=x2−1=(x−0)2−1.
x f(x)
48. Graph f(x)=−2x2+3=−2(x−0)2−(−3).
1 1
49. We will use the point-slope equation.
50. First we find the slope of the given line.
51. Familiarize. Let r= the speed of the boat in still water,
in km/h. We organize the information in a table.
r−4and t2=60
r+4.
Since the total time of the trip is 8 hr, we have
52. Familiarize. Let land wrepresent the length and width
l=56 −2w
−b
2a=−28
2(−1) = 14.
The second coordinate of the vertex is
53. Familiarize. Let h= the length of a side of the hexagon
Cumulative Review Chapters 1 – 11 341
54. Familiarize. Let n= the number of hours it takes the
faster pipe to fill the tank working alone. Then n+4=
55. f(x)=5x2−20x+15
56. f(x)=x4−6x2−16
57. 2x+1
x=3+7
2x+1
x
Now we solve the other equation.
4(2x+1)=x(110 + 14√61) Multiplying by 4x
58. a3
8+8b3
729 =a
23
+2b
93