Try It
2.1 The Rectangular Coordinate Systems and Graphs
1.
xx | y=12x+2y=12x+2 | (x,y)(x,y) |
−2−2 | y=12(−2)+2=1y=12(−2)+2=1 | (−2,1)(−2,1) |
−1−1 | y=12(−1)+2=32y=12(−1)+2=32 | (−1,32)(−1,32) |
00 | y=12(0)+2=2y=12(0)+2=2 | (0,2)(0,2) |
11 | y=12(1)+2=52y=12(1)+2=52 | (1,52)(1,52) |
22 | y=12(2)+2=3y=12(2)+2=3 | (2,3)(2,3) |
2.
x-intercept is (4,0);(4,0); y-intercept is (0,3).(0,3).
4.
(−5,52) (−5,52)
2.2 Linear Equations in One Variable
5.
x=−717. x=−717. Excluded values are x=−12x=−12 and x=−13.x=−13.
10.
Horizontal line: y=2y=2
11.
Parallel lines: equations are written in slope-intercept form.
2.3 Models and Applications
2.
C=2.5x +3,650C=2.5x+3,650
4.
L=37L=37 cm, W=18W=18 cm
2.4 Complex Numbers
1.
−24=0+2i6 −24=0+2i6
3.
(3−4i)−(2+5i)=1−9i(3−4i)−(2+5i)=1−9i
2.5 Quadratic Equations
1.
(x−6)(x+1)=0; x=6, x=−1(x−6)(x+1)=0;x=6, x=−1
2.
(x−7)(x+3) =0, (x−7)(x+3)=0, x=7,x=7 , x=−3.x=−3.
3.
(x+5)(x−5)=0, (x+5)(x−5)=0, x=−5,x=−5, x=5.x=5.
4.
(3x+2)(4x+1)= 0,(3x+2)(4x+1)=0, x=−23,x=−23, x=−14x=−14
5.
x=0,x=−10,x=−1x=0,x=−10,x=−1
8.
x=−23,x=−23, x=13x=13
2.6 Other Types of Equations
4.
0,0, 12,12, −12−12
5.
1;1; extraneous solution −29−29
6.
−2;−2; extraneous solution −1−1
10.
−1,−1, 00 is not a solution.
2.7 Linear Inequalities and Absolute Value Inequalities
2.
(−∞,−2)∪[ 3,∞)(−∞,−2)∪[3,∞)
6.
[−314,∞) [−314,∞)
7.
6<x≤9or(6,9]6<x≤9or(6,9 ]
8.
(−18,12)(−18,12)
10.
k≤1k≤ 1 or k≥7;k≥7; in interval notation, this would be (−∞,1]∪ [7,∞).(−∞,1]∪[7,∞).
2.1 Section Exercises
1.
Answers may vary. Yes. It is possible for a point to be on the x-axis or on the y-axis and therefore is considered to NOT be in one of the quadrants.
3.
The y-intercept is the point where the graph crosses the y-axis.
5.
The x-intercept is (2,0)(2,0) and the y-intercept is (0,6). (0,6).
7.
The x-intercept is (2,0) (2,0) and the y-intercept is (0,−3).(0,−3 ).
9.
The x-intercept is (3,0)(3,0) and the y-intercept is (0 ,98).(0,98).
23.
(3,−32) (3,−32)
31.
not collinear
33.
A: (−3,2) ,B: (1,3),C: (4,0)A: (−3,2),B: (1,3),C: (4,0)
35.
xx | yy |
−3−3 | 1 |
0 | 2 |
3 | 3 |
6 | 4 |
53.
x=−1.667y=0x=−1.667y=0
55.
15−11.2=3.8mi15−11.2=3.8mi shorter
59.
Midpoint of each diagonal is the same point (2,–2)(2,–2). Note this is a characteristic of rectangles, but not other quadrilaterals.
2.2 Section Exercises
1.
It means they have the same slope.
3.
The exponent of the xx variable is 1. It is called a first-degree equation.
5.
If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).
17.
x≠−4; x≠−4; x=−3x=−3
19.
x≠1;x≠1; when we solve this we get x=1,x=1, which is excluded, therefore NO solution
21.
x≠0;x≠0; x=−52x=−52
23.
y=−45x+145 y=−45x+145
25.
y=−34x+2y=−34x+2
27.
y=12x+52y=12x+52
37.
Parallel
39.
Perpendicular
45.
m1=−13, m2=3;Perpendicular.m1=−13,m2=3;Perpendicular.
47.
y=0.245x−45.662. y=0.245x−45.662. Answers may vary. ymin=−50,ymax=−40ymin=−50,ymax=−40
49.
y=−2.333x+6.667.y =−2.333x+6.667. Answers may vary. ymin=−10, ymax=10ymin=−10, ymax=10
51.
y=−ABx+CBy=−ABx+CB
53.
The slope for (−1,1)to (0,4)is 3.The slope for (−1,1)to (2,0)is −13.The slope for (2,0)to (3,3)is 3. The slope for (0,4)to (3,3)is −13.The slope for (−1,1)to (0,4)is 3.The slope for (−1,1)to (2,0)is −13.The slope for (2,0)to (3,3)is 3.The slope for (0,4)to (3,3)is −13.
Yes they are perpendicular.
2.3 Section Exercises
1.
Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.
7.
Ann: 23;23; Beth: 4646
21.
She traveled for 2 h at 20 mi/h, or 40 miles.
23.
$5,000 at 8% and $15,000 at 12%
25.
B=100+.05xB=100+ .05x
33.
W= P−2L2=58−2(15)2=14W=P−2L2=58−2(15)2=14
35.
f=pqp+q=8(13)8+13=10421f=pqp +q=8(13)8+13=10421
39.
h=2Ab1+b2 h=2Ab1+b2
41.
length = 360 ft; width = 160 ft
45.
A=88in.2 A=88in.2
49.
h=Vπr2h=Vπr2
2.4 Section Exercises
1.
Add the real parts together and the imaginary parts together.
3.
Possible answer: ii times ii equals -1, which is not imaginary.
9.
−2329+1529i−2329+1529i
33.
25+115i 25+115i
45.
(32+12i) 6=−1(32+12i)6=−1
55.
92−92i92 −92i
2.5 Section Exercises
1.
It is a second-degree equation (the highest variable exponent is 2).
3.
We want to take advantage of the zero property of multiplication in the fact that if a⋅b=0a⋅b=0 then it must follow that each factor separately offers a solution to the product being zero: a=0 orb=0.a=0orb=0.
5.
One, when no linear term is present (no x term), such as x2=16.x2=16. Two, when the equation is already in the form (ax+b)2=d.( ax+b)2=d.
9.
x=−52,x=−52, x=−13 x=−13
13.
x=−32,x=−32, x=32 x=32
17.
x=0,x=0, x=−37x=−37
25.
x=−2, x=−2, x=11x=11
29.
z=23,z=23 , z=−12z=−12
31.
x=3±174 x=3±174
39.
x=−1±172x=−1±172
41.
x=5±136x=5±136
43.
x=−1±178x=−1±178
45.
x≈0.131x≈0.131 and x≈2.535x≈2.535
47.
x≈−6.7x≈−6.7 and x≈1.7x≈1.7
49.
ax2+bx+c=0x2+bax=−cax2+bax+b2 4a2=−ca+b4a2(x+b2a)2 =b2−4ac4a2x+b2a=±b2−4a c4a2x=−b±b2−4ac2a ax2+bx+c=0x2+bax=−cax2+bax+b24a2=−ca+b4a2(x+b2a)2=b2−4ac4a2x+b 2a=±b2−4ac4a2x=−b±b2 −4ac2a
51.
x(x+10)=119;x(x+10)=119; 7 ft. and 17 ft.
55.
The quadratic equation would be (100x−0.5x2 )−(60x+300)=300.(100x−0.5x2)−(60x+300)=300. The two values of xx are 20 and 60.
2.6 Section Exercises
1.
This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.
3.
He or she is probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27.−27.
5.
A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.
15.
y=0, 32, −32y=0, 32, −32
19.
x=25,±3ix= 25,±3i
31.
x=−54,74x=−54,74
37.
x=1,−1,3,-3x =1,−1,3,-3
45.
x=4,6,−6,−8x=4,6,−6,−8
2.7 Section Exercises
1.
When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.
5.
We start by finding the x-intercept, or where the function = 0. Once we have that point, which is (3,0),(3,0), we graph to the right the straight line graph y=x−3, y=x−3, and then when we draw it to the left we plot positive y values, taking the absolute value of them.
7.
(−∞,34] (−∞,34]
9.
[−132,∞) [−132,∞)
13.
(−∞,−373 ](−∞,−373]
15.
All real numbers (−∞,∞) (−∞,∞)
17.
(−∞,−103 )∪(4,∞)(−∞,−103)∪(4,∞)
19.
(−∞,−4] ∪[8,+∞)(−∞,−4]∪[8,+∞)
27.
[−10,12][−10,12]
29.
x>− 6andx>−2Take the intersection of two sets.x>−2,(−2,+∞)x>−6andx>−2Take the intersection of two sets.x>−2,(−2,+∞)
31.
x<− 3orx≥1Take the union of the two sets.(−∞,−3)∪ [1,∞)x<−3orx≥1 Take the union of the two sets.(−∞,−3)∪[1,∞)
33.
(−∞,−1)∪(3,∞)(−∞,−1)∪(3,∞)
35.
[−11 ,−3][−11,−3]
37.
It is never less than zero. No solution.
39.
Where the blue line is above the orange line; point of intersection is x=−3. x=−3.
(−∞,−3)(−∞,−3)
41.
Where the blue line is above the orange line; always. All real numbers.
(−∞,−∞)(−∞,−∞)
47.
{x|x<6} {x|x<6}
49.
{x|−3≤x<5 }{x|−3≤x<5}
55.
Where the blue is below the orange; always. All real numbers. (−∞ ,+∞).(−∞,+∞).
57.
Where the blue is below the orange; (1,7).(1,7).
63.
80≤T≤120 1,600≤20T≤2,40080≤T≤1201,600≤20T≤2,400
[1,600,2,400][1,600,2,400]
Review Exercises
1.
x-intercept: (3,0); (3,0); y-intercept: (0,−4)(0,−4)
9.
midpoint is (2, 232)(2,232)
19.
y=16x+43y= 16x+43
21.
y=23x+6y=23x+6
27.
x=−34±i474 x=−34±i474
29.
horizontal component −2;−2; vertical component −1−1
47.
x=−1±54x=−1±54
49.
x=25,−13 x=25,−13
59.
x=112,−172x=112,−172
63.
[−103,2] [−103,2]
67.
(−43,15)(−43,15)
69.
Where the blue is below the orange line; point of intersection is x=3.5.x= 3.5.
(3.5,∞)(3.5,∞)
Practice Test
1.
y=32x+ 2y=32x+2
3.
(0,−3)(0,−3 ) (4,0)(4,0)
9.
x≠−4,2;x≠−4,2; x=−52,1x=−52,1
15.
y=−59x−29y =−59x−29
17.
y=52x−4y=52x−4
21.
513−1413i 513−1413i
25.
x=12±22x=12±22
29.
x=12,2,−2 x=12,2,−2