Download NONLINEAR FUNCTIONS 2.1 Properties of Functions

Chapter 2
NONLINEAR FUNCTIONS
9. i/ = 2x + 3
2.1
Properties of Functions
1. The x-value of 82 corresponds to two y-values, 93
and 14. In a function, each value of x must corre
spond to exactly one value of y.
X
-2
-l
0
1
2
3
y
-l
l
3
5
7
9
Pairs: (-2,-1), (-1,1), (0,3),
(1,5), (2,7), (3,9)
The rule is not a function.
Range: {-1, 1, 3, 5, 7, 9}
2. Each x-value corresponds to exactly one ?/-value.
The rule is a function.
9-
•
3. Each x-value corresponds to exactly one t/-value.
The rule is a function.
.I--
I I 1 I I I I
4. 9 corresponds to 3 and -3, 4 corresponds to 2 and
—2, and 1 corresponds to —1 and 1.
I I I I I I
3
The rule is not a function.
5. y = x3 + 2
10. y = -3x + 9
Each x-value corresponds to exactly one y-value.
X
-2
-1
0
1
2
3
The rule is a function.
y
15
12
9
6
3
0
Pairs:
6. y= Jx~
Each x-value corresponds to exactly one y-value.
(-2,15), (-1,12), (0,9),
(1,6), (2,3), (3,0)
Range: {0,3,0,9,12,15}
The rule is a function.
(-2,15),
14
7. x = \y\
Each value of x (except 0) corresponds to two yvalues.
6
The rule is not a function.
2
I
I
I
(3,0)
I,
MM
8. x = t/2 + 4
Solve the rule for y.
y2 = x - 4 or y = ±\/x - 4
11. 2y-x = 5
2y = 5 + x
V=
Each value of x (greater than 4) corresponds to
y
1
5
—x H—
2
two y-values
X
y = \/x —4 and y = —\Jx —4.
The rule is not a function.
94
2
-2
3
y
2
-1
0
1
2
5
2
3
2
V
2
3
4
Pairs:(-2,§), (-1,2), (0,§),(!, 3), (2,J), (3,4)
Section 2.1
95
Properties of Functions
14. y = (x- 2)(x + 2)
Range: {§, 2, § 3, j, 4}
X
::
y
.0,4)
-2
-l
0
1
2
3
0
-3
-4
-3
0
5
Pairs: (-2,0), (-1,-3), (0,-4),
(1,-3), (2,0), (3,5)
Range: {-4,-3,0,5}
I I I I I I
.(3,5)
2--
(-2.0)
I II ♦ I,
I M
X
-2
-1
0
1
2
3
y
4
1
0
1
4
9
12. 6x-y = -1
—j/ = —6x —1
y = Qx + 1
X
-2
-l
0
1
2
3
y
-li
-5
1
7
13
19
15. y = x2
Pairs: (-2,-11), (-1,-5), (0,1),
(1,7), (2,13), (3,19)
Range: {-11,-5,1,7,13,19}
Pairs: (-2,4), (-1,1), (0,0),
(1,1), (2,4), (3,9)
y
,(3,19)
20 +
I I
Range: {0, 1, 4, 9}
• (3,9)
I
I
I
I
'.. I
(-2.4).'
(-2,-11)
-I I I I I I I
I 1 I I I I I
3
13. y = x(x + 2)
X
-2
-l
0
l
2
3
y
0
-l
0
3
8
15
Pairs: (-2,0), (-1,-1), (0,0),
(1,3), (2,8), (3,15)
16. y = -4x2
X
-2
-1
0
1
2
3
y
-16
-4
0
-4
-16
-36
Pairs:
(-2,-16), (-1,-4), (0,0),
(1,-4), (2,-16), (3,-36)
Range: {-36,-16,-4,0}
Range: {-1, 0, 3, 8, 15}
.V
I
.(3, 15)
I
I
I
I
I
•
I
I
I
3
10-
(-2,-16).
•
-20-•
(-2,0)
II
f
"
i
ii
0
i
i
2
i
i
i
•*
-35
. (3, -36)
96
Chapter 2
17. y =
20. y =
x + 3
2x+l
x + 3
X
-2
-l
0
1
2
3
y
l
i
2
1
3
i
l
5
l
6
4
Pairs: (-2,1), (-1,I), (0, |), (1,1), (2, J), (3,1)
Range: {1, 1, 1, J, 1, 1}
X
-2
-1
0
1
2
y
-3
1
2
1
3
4
1
3
3
1
6
Pairs: (-2,-3), (-!,-£),
(0, J)
23-"
(l,f), (2,1), (3, |)
Range: {-3, -| 1 f, 1,1}
(-2,1)
7 T T •
H—h
18. y =
NONLINEAR FUNCTIONS
2--
H—I—h
-2
H—I—h
x + 4
X
-2
-l
/y
-l
2
3
0
1
1
2
2
5
2
3
l
2
7
3
Pairs: (-2, -1), (-1, -f), (0,-1),
21. /(x) = 2x
(1,-|), (2,-1), (3, -f)
Range: {-1,-§,-I,-§,-I,-|}
x can take on any value, so the domain is the set
of real numbers, (-co, oo).
22. f(x) = 2x + 3
x can take on any value, so the domain is the set
2
H
1
1-
I
1
of real numbers, which is written (—00,00).
i
23. f(x) = x4
x can take on any value, so the domain is the set
of real numbers, (—00,00).
2x-2
19. y =
x + 4
24. /(x) = (x + 3)2
X
-2
-1
y
-3
4
3
0
1
2
1
2
3
0
l
3
4
7
x can take on any value, so the domain is the set
of real numbers, (—00,00).
Pairs:(-2,-3), (-1,-f), (0,-£),(l,0), (2,1), (3,|)
Range: {-3, -§, -A, 0, 1, f }
25. f{x) = n/I^x2"
For f(x) to be a real number, 4 —x2 > 0.
I
I
I
■♦-+
.M)
Solve 4 - x2 = 0.
(2-x)(2 + x)=0
x = 2
(-2,-3).
or
x =
-2
_3
The numbers form the intervals (—00, —2), (—2,2),
and (2,oo).
Only values in the interval (—2,2) satisfy the in
equality. The domain is [—2,2].
Section 2.1
97
Properties of Functions
26. f(x) = |3x - 6|
31. /(x) = -
x can take on any value, so the domain is the set
of real numbers, (—00,00).
27. f(x) = (x-3)l'2 = >/x~=3
(x - 4)(x + 4)
16
(x-4)-(x+4) > 0, since (x-4)-(x+4) < 0 would
produce a negative radicand and (x—4)(x+4) = 0
would lead to division by zero.
For f(x) to be a real number,
Solve (x - 4) • (x + 4) = 0.
x-3>0
£-4 = 0
or
x = 4
or
x + 4
=0
x = —4
x>3.
Use the values —4 and 4 to divide the number
line into 3 intervals, (-00, -4), (-4,4) and (4,00).
Only the values in the intervals (—00, —4) and
(4,00) satisfy the inequality.
The domain is [3,00).
28. f(x) = (3x+ 5)l/2 = v/3x + 5
The domain is
For f(x) to be a real number,
(-00,-4) U (4,00).
3x + 5 > 0
3x> -5
i(3x) > I(-B)
1-x2
+ 36
duce a zero in the denominator. Also, no choice
3"
In interval notation, the domain is [—|,oo).
29. f(x) =
-"Vx^
x can take on any value. No choice for x will pro
x>-5
-
32. f(x) = -
(l-x)(l + x)
Since division by zero is not defined,
for x will produce a negative number under the
radical. The domain is (—00,00).
33. f(x) = s/x2 - 4x - 5 = y/(x-5)(x+l)
See the method used in Exercise 25.
(l-x)-(l + x)^0.
When(l-x)(l + x)=0,
(x-5)(x + l)>0
when x > 5 and when x < —1.
1 - x = 0
or
x = 1
or
1 + x = 0
The domain is (—00, —1] U [5,00).
x = —1.
34. f(x) = Vl5x2 + x - 2
Thus, x can be any real number except ±1.
The expression under the radical must be nonneg-
The domain is
ative.
(-oo,-l)U(-l,l)U(l,oo).
15x2 + x - 2 > 0
(5x + 2)(3x-l) >0
30. /(x) =
-8
Solve (5x + 2)(3x - 1) = 0.
x2 - 36
In order for /(x) to be a real number, x2 —36
cannot be equal to 0.
5x + 2 = 0
5x = -2
When x2-36 = 0,
x
x2 = 36
x = 6
or
or
x = —6.
Thus, the domain is any real number except 6 or
—6. In interval notation, the domain is
(-00, -6) U (-6,6) U (6,00).
=
3x - 1 = 0
3x = 1
2
1
5
X=3
—
Use these numbers to divide the number line into
3 intervals, (—00, —|), (—§, 5), and (5,00).
Only values intheintervals (—00, —|) and (3,00)
satisfy the inequality. The domain is (—00, —|] U
[*.<»)•
98
Chapter 2
35. /(*) =
v/3x2+2x-l
x/(3x-l)(x+l)
(b)
NONLINEAR FUNCTIONS
'(-J)-(-i)'-«(->
(3 —2)(x + 1) > 0, since the radicand cannot be
=7+2+1
4
negative and the denominator of the function can
= 15
not be zero.
Solve(3-l)(x+l) = 0.
4
(c) f(a) = 3(a)2 - 4(a) + 1
3-1 = 0
or
x + l=0
x =
or
x = -l
£
= 3a2 - 4a + 1
Use the values -1 and ^ to divide the number line
into 3 intervals, (-co,-1), (-1,4) and (|,oo).
Only the values in the intervals (-oo,-|) and
12
8
m£
m
(5,00) satisfy the inequality.
12 - 8m + m2
or
The domain is (—oo, —1) U(|,oo).
nv
(e)
f{x) = 1
3x2 - 4x + 1 = 1
3x2 - 4x = 0
36. J{x) =
3-x
x(3x - 4) = 0
For f(x) to be a real number,
x = 0
x2 > 0
—oo < x < oo
and
3- x > 0
and
x<3
x<3.
The domain is (—oo,3).
37. By reading the graph, the domain is all numbers
greater than or equal to —5 and less than 4. The
range is all numbers greater than or equal to -2
and less than or equal to 6.
Domain: [—5,4); range: [—2,6]
or
42. /(x) = (x + 3)(x-4)
(a) /(4) = (4 + 3)(4-4) = (7)(0) = 0
M-iM-H(-H
-G)H)~?
(c) fia) = [(a) + 3][(o) - 4] = (a + 3)(a - 4)
M!)- CHGH
_ /2_+_3m\ /2-4m\
_ (2 + 3m)(2 - 4m)
Range: [0, oo)
39. By reading the graph, x can take on any value,
but y is less than or equal to 12.
Domain: (—00,00); range: (—00,12]
40. By reading the graph, both x and y can take on
any values.
Domain: (—00,00)
4
x = o
38. By reading the graph, the domain is all numbers
greater than or equal to -5. The range is all num
bers greater than or equal to 0.
Domain: [-5, oo)
,
= —-- + 1
in2
2(2 + 3m)(1 - 2m)
or
11V
(e)
fix) = 1
(x + 3)(x-4) = l
x2 - x - 12 = 1
x2 - x - 13 = 0
Range: (—00,00)
1 ± s/1 + 52
2
41. f(x) = 3x2 - 4x + 1
l±\/53
x
=
(a) /(4)=3(4)2-4(4) + l
= 48-16 + 1
= 33
x « -3.140 or x& 4.140
Section 2.1
99
Properties of Functions
2x+l
43. fix) =
(e)
x-2
3x + 2
2(4) + 1
(a) /(4) =
f{x) = 1
2x-4
4-2
2
3x + 2 = 2x - 4
2(-i) + l
(-0 =
(b)/ -;
x = —6
-A2 -2
z
45. The domain is all real numbers between the end-
-1 + 1
—
points of the curve, or [-2,4].
The range is all real numbers between the mini
mum and maximum values of the function or [0,4].
5
2
4 =o
2(a) + 1
(c) /(a) =
(a) - 2
(a) fi-2) = 0
2a + 1
a- 2
(b) /(0) = 4
(c)/g)=3
2(i) + l
Mi)
_4
(d) From the graph, /(x) = 1 when x = -1.5, 1.5,
i_ m
T71
m
2
or 2.5.
2m
46. The domain is all real numbers between the end
points of the curve, or [—2,4].
2-2m
The range is all real numbers between the mini
4 + m
mum and maximum valuesof the function or [0,5].
m
m
2 — 2m
(a)/(-2) = 5
4 + m
2-2m
6)
(d) From the graph, if fix) = 1, x = -0.2, 0.5, 1.2, or
(c)/
fix) = 1
(e)
2x + l
= 1
2.8.
x-2
2x + 1 = x - 2
47. The domain is all real numbers between the end-
points of the curve, or [—2,4].
x = -3
44. fix) =
(b)/(0) = 0
3x + 2
The range is all real numbers between the mini
2x-4
mum and maximum values of the function or [—3,2].
3(4) + 2
12 + 2
14
8-4
4
(a>"4> = 2(4)-4
<b> /(-*)-
3(-i)+2
2(4)_
-5
(c) /(a) =
3(a) + 2
2(a) - 4
Ml)-
-f+2
2
-5
4
m
W/6)-i
__J_
(d) From the graph, fix) = 1 when x = 2.5.
points of the curve, or [—2,4].
2a
The range is all real numbers between the mini
mum and maximum values of the function, or in
6+2m
Am
4—4m
m
m
4 — 4m
4m
(a) fi-2) = 3
(b) /(0) = 3
m
m
4 -
this case, {3}.
2m
6 + 2m
10
48. The domain is all real numbers between the end
3a+ 2
6 +2m
(b) /(0) = -2
-1-4
3(£)+2
2(4)
+
(a) fi-2) = -3
(c)/(i)=3
3 + m
2 -
2m
(d) From the graph, /(x) is 1 nowhere.
100
Chapter 2
49. fix) = 6x2 - 2
59. A vertical line drawn anywhere through the graph
will intersect the graph in only one place. The
/(* + 1) = 6(t + I)2 - 2
graph represents a function.
= 6(*2 + 21 + 1) - 2
= 6t2 + 12* + 6 - 2
= 6*2 + 124 + 4
50.
NONLINEAR FUNCTIONS
60. A vertical line is not a function since the one x-
value in the domain corresponds to more than one,
in fact, infinitely many y-values. The graph does
fix) = 6x2 - 2
/(2 - r) = 6(2 - r)2 - 2
= 6(4 - 4r + r2) - 2
= 24 - 24r + 6r2 - 2
= 6r2 - 24r + 22
not represent a function.
61. f{x) = 2x + 1
(a) f{x + h) = 2(x + h) + 1
51. gir + h)
= (r + h)2 - 2(r + /i) + 5
= 2x + 2/i + 1
= r2 + 2/*r + /i2 - 2r - 2h + 5
(b) fix+ h)-fix)
52.a(2-p)
2x + 2/j + 1
= (*-p)2-2(z-p) + 5
= z2 - 2-zp + p2 - 2z + 2p + 5
2.T+1
= 2x + 2/i + 1 •- 2 x - 1
= 2h
9
g2
6
a
,_, fix+ h)-fix)
r
{}
= -2 --+5
h
2.r+2/i+l
2a:+l
9 - 6g + W
h
or
2x + 2h + 1 -- 2 x - 1
2
h
"•(-?)-(-!)-(->
2h
h
25
10 r
= -^ + — +5
_ 25
z2
= 2
lOz
5z2
z2
z2
62.
f(x) = x2 - 3
25 + 102 + 5z2
55. A vertical line drawn anywhere through the graph
will intersect the graph in only one place. The
(a) fix + h) = (x + h)2 - 3
= (x2 + 2xh + h2) - 3
= x2 + 2x/i + h2 - 3
graph represents a function.
56. A vertical line drawn anywhere through the graph
will intersect the graph in only one place. The
(b) fix+ h)-fix)
= (x2 + 2xh + h2 - 3)
-(x2-3)
graph represents a function.
= x2 + 2xh + h2 - 3 - x2 + 3
= 2xh + h2
57. A vertical line drawn through the graph may in
tersect the graph in two places. The graph does
not represent a function.
58. A vertical line drawn through the graph may inter
sect the graph in two or more places. The graph
does not represent a function.
(c)
f{x + h) - fix)
2xh + It
h
h
2x + h
h
= 2x + h
Section 2.1
101
Properties of Functions
65. J{x) = i
63. fix) = 2x2 - Ax - 5
(a)/(x + ft) =
(a) f(x + h)
= 2(x + ft)2-4(x + ft)-5
= 2(x2 + 2hx+ ft2) - Ax -Ah-5
(b) /(x + ft)-/(x)
__L
= 2x2 + Ahx + 2h2 -Ax-Ah-5
1
x + h
x
—(x\ *
—Ix/x + ft
(b) f(x + h)-f(x)
~
- {2x2 - Ax - 5)
x(x + ft)
(c)
h
x(x + ft)
-ft
= 2a:2 + 4ftx + 2h2 - Ax - 4ft - 5
- 2x2 + Ax + 5
= Ahx + 2h2 - Ah
/(x + ft)-/(x)
I fx + h\
x\x + hj
x - (x + ft)
= 2x2 + Ahx + 2h2 -Ax-Ah-b
(c)
1
x + ft
/(.T + ft)-/(x)
ft
_J
i
x+h
x
1 lx\ _ i fx+h\
x+h \x) x yx+hJ
Ahx + 2h2 - Ah
h
ft
h{Ax + 2h - 4)
x-(x+h)
(x+h)x
h
ft
= Ax + 2h - A
1
x — X — ft
ft
(re + h)x
1
ft
64. f{x) = -Ax2 + Zx + 2
'
-ft
"
(re + ft)x
-1
x(x + ft)
(a) /(x + ft)
= -4(x + ft)2 + 3(x + ft) + 2
= -4(x2 + 2ftx + h2) + Zx + 3ft + 2
= -4x2 - 8ftx - 4ft2 + 3x + 3ft + 2
66. f{x) = —g
(a) /(x + ft) = -
1
(x + ft)2
1
re2 + 2xft + ft2
(b) f(x + h)-f(x)
= -Ax2 - 8hx - Ah2 + Zx + 3ft + 2
(b) /(x + ft)-/(x) = -
1
(-A)
re2 + 2xft + ft2
-(-4x2+3x + 2)
1
1
= -4x2 - 8ftx - Ah2 + Zx + 3ft + 2
+ 4x2 - 3x - 2
= -8ftx - Ah2 + 3ft
x2 + 2xft + ft2
+ ^7
re2
x2(x2+2xft + ft2)
(x2 + 2xft + ft2)
(c)
/(x + ft) - /(x)
h
-Shx - Ah2 + Zh
~
+
x2(x2 + 2xft + ft2)
h
-x2 + x2 + 2xft + ft2
h(-8x -4/i + 3)
x2(x2 + 2xft + ft2)
2xft + ft2
= -8x -4/i + 3
x2(x2 + 2xft + ft2)
102
Chapter 2
h(2x+h)
/c\ fiX + ty ~ fix) = g!*(;c*+2gft+ft*)
73.
fix) =
ft
2x + ft
fi-x) =
x2(x2 + 2xft + ft2)
NONLINEAR FUNCTIONS
x
x2 - 9
—x
(-x)2 - 9
x
X'
67.
fix) = Zx
= -fix)
fi-x) = 3(-x)
The function is odd.
= -(3a;)
= -/(*)
The function is odd.
68.
fix) = 5x
/(-*) = 5(-x)
= -(5x)
= -fix)
The function is odd.
9
74.
J{x) = \x-2\
fi-x) = \-x-2\
= |-(x + 2)|
= |x + 2|
fi-x) does not equal either /(x) or /(-x). The
function is neither even nor odd.
75. (a) The independent variable is the years.
(b) The dependent variable is the number of In
ternet users.
69.
fix) = 2x2
/(-x) = 2(-x)2
= 2x2
= /(*)
The function is even.
70.
(c) /(2003) = 719 million users.
(d) The domain is 1995 < x < 2006.
The range is 16,000,000 < rj < 1,043,000,000.
76. If x is a whole number of days, the cost of renting
= x2 - 3
a saw in dollars is 5(x) = 28x + 8. For x in whole
days and a fraction of a day, substitute the next
whole number for x in 28x + 8, because a fraction
of a day is charged as a whole day.
= fix)
(a) 5(1) =5(1) = 28(1) +8 = 36
fix) = x2 - 3
/(-x) = (-x)2-3
The function is even.
The cost is $36.
(b) 5(1) = 28(1)+8 = 36
71.
F(-r) —
/{X)-x2 +A
—
Jif(-r\
X>-i-x)2+4
1
x2 + 4
= fix)
The cost is $36.
(c) 5 (1|) = 5(2) = 28(2) + 8 = 56 + 8 = 64
The cost is $64.
(d) 5 (3^) = 5(4) = 28(4) + 8 = 112 + 8 = 120
The cost is $120.
The function is even.
(e) 5(4) = 28(4) + 8 = 112 + 8 = 120
72.
fix) = X3 + X
fi-x) = i-xf + (-X)
= —x3 — X
The cost is $120.
(f) 5 (4-jL) = 5(5) = 28(5) + 8 = 140 + 8 =
= -(x3+x)
The cost is $148.
= -fix)
(g) 5 (4^) = 5(5) = 28(5) + 8 = 140 + 8 =
The function is odd.
The cost is $148.
148
148
Section 2.1
103
Properties of Functions
(h) To continue the graph, continue the horizon
79. (a)
(i)
y = /(5) = 19.7(5)0-753
(ii)
y = / (25) = 19.7(25)0753
« 66 kcal/day
tal bars up and to the right.
(i) The independent variable is x, the number of
« 222 kcal/day
full and partial days.
(j) The dependent variable is S, the cost of renting
a
saw.
(b) Since 1 pound equals 0.454 kg, then
x = g(z) = 0.4542 is the number of kilograms
equal in weight to z pounds.
77. If a: is a whole number of days, the cost of renting
a car is given by
C(x) = 54.r + 44.
80. (a)(i) By the given function /, a muskrat weigh
ing 800 g expends
/(800) = 0.01(800)088
For x in whole days plus a fraction of a day, sub
« 3.6, or approximately
stitute the next whole number for x in 54x + 44
because a fraction of a day is charged as a whole
day.
3.6 kcal/km when swimming at the surface of the
water.
(a)C(|)=C(l)
(ii) A sea otter weighing 20,000 g expends
= 54(1) + 44
= $98
/(20,000) = 0.01(20,000)088
(b) C(±)=C(1)
« 61, or approximately
61 kcal/km when swimming at the surface of the
(c) C(l) = $98
water.
(d) C(1|)=C(2)
= 54(2)+44
(b) If z is the number of kilogram of an amimal's
= $152
weight, then x = g(z) = lOOOz is the number of
grams since 1 kilogram equals 1000 grams.
(e) C(2.4) = C(3)
= 54(3) -I- 44
81. (a) In the graph, the curves representing wood
and coal intersect approximately at the point (1880,50).
= $206
(f)
So, in 1880 use of wood and coal were both about
50% of the global energy consumption.
314260
206
(b) In the graph, the curves representing oil and
coal intersect approximately at the point (1965,35).
152+
98
-I
1
1
1
h
12
3
4
5
Number of days
So, in 1965 use of oil and coal were both about
35% of the global energy consumption.
82. (a) P = 2L + 2W
(g) Yes, C is a function.
(h) No, C is not a linear function.
However, LW = 500, so L = -777-.
W
78. (a) The curve in the graph crosses the point with
P(W) = 2
rr-coordinate 17:37 and y-coordinate of approxi
mately 140. So, at time 17 hours, 37 minutes the
whale reaches a depth of about 140 m.
/500\
+ 2W
P(W) =^ +2W
(b) The curve in the graph crosses the point with
x-coordinate 17:39 and y-coordinate of approxi
mately 240. So, at time 17 hours, 39 minutes the
whale reaches a depth of about 240 m.
(b) Since L = ^a, W ^ 0 but W could be any
positive value. Therefore, the domain of P is 0 <
W < 00.
104
Chapter 2
(c)
(Pg(l,00Q/n-)"+2wj
300
NONLINEAR FUNCTIONS
6. The graph of y = (x+3)2+2 is the graph of y = x2
translated 3 units to the left and 2 units upward.
This is graph b.
0^
J 100
7. The graph of y = -(3 - x)2 4- 2 is the same as the
graph of y = -(x - 3)2 + 2. This is the graph of
y = x2 reflected in the .T-axis, translated 3 units
to the right, and translated 2 units upward.
This is graph c.
83. (a) Let w = the width of the field;
/ = the length.
The perimeter of the field is 6000 ft, so
2/ + 2w = 6000
8. The graph of y = -(x + 3)2 + 2 is the graph of
y = x2 reflected in the .-c-axis, translated three
units to the left, and two units upward.
This is graph e.
l + w = 3000
I = 3000 - w.
Thus, the area of the field is given by
9. y = x2 + 5x + 6
y = {x + 3)(x + 2)
Set y = 0 to find the ^-intercepts.
A = hu
0 = (.r + 3)(.r + 2)
A = (3000 - w)w.
x = -3, x = -2
(b) Since / = 3000 —w and w cannot be negative,
0 < w < 3000.
The domain of A is 0 < w < 3000.
(c)
The .^-intercepts are —3 and —2.
Set x = 0 to find the y-intercept.
(>4 =(3.000-ii'fr)
y = 02 + 5(0) + 6
y = 6
2,250.000
The {/-intercept is 6.
The .T-coordinate of the vertex is
-b_ -5 __5
x
0^.
!/ 3,000
=
2a. ~
2
2'
Substitute to find the {/-coordinate.
2.2
Quadratic Functions;
/ 5%2 e/ 5,
25
25 „
y=(--)2
+S(--) +G=T-+6= -41
Translation and Reflection
The vertex is (—-, —-).
3. The graph of y = x2 —3 is the graph of y = x2
translated 3 units downward.
The axis is x = —|, the vertical line through the
vertex.
This is graph d.
4. The graph of y = (x - 3)2 is the graph of y = x2
translated 3 units to the right.
This is graph f.
5. The graph of y = (x - 3)2 + 2 is the graph of
y = x2 translated 3 units to the right and 2 units
upward.
This is graph a.
y - x2+ 5.t + 6