Download 1 + (~)2 - NUAMES Mathematics

gect«uu 1-5, 1-6, 1-7, ~
1-8
Section 1-5 introduces techniques for simplifying complex expressions.
Section 1-6 reviews various methods for solving quadratic equations.
Sections 1-7 and 1-8 explore quadratic functions and models algebraically
and graphically.
KEY TERMS
EXflMPLE/ILWSTRflTION
Imaginary unit i (p. 26)
i = ~andi2
=-1
Complex number (p. 27)
any number of the form a + bi, where a and b are real numbers
and i is the imaginary unit (a is the real part and b is the
imaginary part of a + bi.)
-{5
L---..J
real part, imaginary part 0
0.25i
L-J
imaginary part, real part 0
1_ 8i
}\
real part
Imaginary number (p. 27)
a complex number a + bi with b
*-
Pure imaginary number (p. 27)
a complex number with a = 0 and b
2i-{2,9 + 4i
0
*-
imaginary part
-i{5,2.6i
0
Complex conjugates (p. 27)
two complex numbers z = a + bi and Z = a - bi
Quadratic equation (p. 30)
any equation that can be written in the standardform
ax2 + bx + c = 0, where a*-O (A root or solution of a
quadratic equation is a value of the variable that satisfies the
equation.)
Completing the square (p. 30)
the method of transforming a quadratic equation so that one side
is a perfect square trinomial
x 2 + x - 2 = 0 has roots 1 and
- 2 since 1 2 + 1 - 2 = 0 and
(_2)2 + (-2) - 2 = O.
x2 + 4x + 1
x2 + 4x
x2 + 4x + (~) 2
=
=
=
0
-1
-1 +
(x
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ADVANCED MATHEMATICS
+ 2)2
=
(~)2
3
Student Resource Guide
5
Discriminant (p. 31)
the value b2
-
4ac for a quadratic equation ax2 + bx +
C
3x2 + x - 2
= 0
=
=
ax2
0
+ bx + c
0
a = 3,b = 1,c = -2
b2
4ac = 12 - 4(3)(-2)
-
=
Quadratic function (p. 37)
a function of the formf(x)
graph is a parabola
= ax2 + bx + c, a
25
* 0, whose
y = 2x2 - 3x- 5
Quadratic model (p. 44)
a quadratic function that models a real-world situation
If a certain type of carpeting costs
$40/yd 2, then the cost of
carpeting a square room y yards
on a side is C(y) = 40y2.
O"DERSTA"DI"G THE MAl" IDEAS
Complex numbers
• To simplify an expression that contains ---~
the square root of a negative number, begin by
writing the expression in terms of i.
-V - 27 -
=
• To add or subtract complex numbers, group the real parts and the imaginary parts.
70 - (-2 + 0
=
~
• To simplify an expression with a complex
number in the denominator, multiply the
numerator and denominator by the complex
conjugate of the denominator.
C
(5 -
(3{3 - {3)i
2i{3
=
5 - 7i + 2 - i
= 7 - 8i
1
10 + 13i - 3(-1)
=
1 + i-V3
1
1+
l--D
1-i{3
= 1-3(-1)
=
13 + 13i
1-i{3
1-i{3
=
1- 3i2
1-i{3
{3.
1
= - --1
4
4
= 0
• Use factoring if a, b, and c are integers and b2
is a perfect square.
-
4ac
2x2 - 7x - 4 = 0
+ 1)(x - 4)
0
2x+ 1 =00rx-4=0
=
(2x
1
2
x = --or
6 ADVANCED MATHEMATICS
=
(2 + 3i)(5 - i) = 10 - 2i + 15i - 3i2
• To multiply two complex numbers use FOIL (first, outer, inner, last) and substitute
-lfori2.
Solving a quadratic equation: ax2 + bx +
-V-3 = i{27 - r{3
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x = 4
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• Solve by completing the square if a
x2 + 4x + 1
(x + 2)2
x + 2
x
=
=
°
=
1 and b is even.
3 (See the example in Key Terms.)
= ±{3
= -2 +
(Take the square root of both sides.)
=
{3 or x
.
• Use the quadratic formula x
=
-2 - {3
-b±--.Jb2-4ac
2a
otherwise.
(See Example 3 on text page 31.)
Graphing a quadratic function
• y = ax2 + bx + c (See Example 1 on text page 38.)
If a > 0, the graph is U-shaped.
If a < 0, the graph is n-shaped.
The point (0, c) is on the graph.
-b + --.J b2 - 4ac)
2a
,0
and
(
(-b - -Vr-b2-_-4a-c)
2a
,0 are on the graph.
The equation of the axis of symmetry is
The vertex has x-coordinate • y =
If a
If a
The
x = - ;a .
;a .
a(x - h)2 + k ------------_~
> 0, the graph is U-shaped.
< 0, the graph is n-shaped.
vertex is (h, k) and the axis is x
=
x
h.
Quadratic models
If you have a quadratic modelf(x) = ax2 + bx + c,
you can use the model to predict data values (see
Example 2 on text page 45) or to maximize or
minimize the function (evaluate fwhen
y
= -2(x + 1)2 -
3
x = - ;a) .
CHECKltlG THE MAlti IDEAS
1. (a + bi)2
A.
c.
a2 a2 +
=
_?_
B. a2 + b2
D. a 2 + 2abi + b2
b2
2abi - b2
4
2. Simplify (a) _~
and (b) -1 . + i.
,,-20
-1
3. Solve x2 - 2x + 5
4. Sketch the graph of y
=
°
by using two different methods. Show your work.
= -~
x2 - 2x + 6. Label the vertex, axis of
symmetry, and intercepts of the parabola.
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ADVANCED MATHEMATICS
Student Resource Guide
7
5. The table below shows the area A of a sector of
a circle with perimeter 12 and radius r.
=
~
a. Find a quadratic function that models this data. (See Example 1 on
text page 44.)
b. Find the maximum possible area.
c. Critical Thinking Find the domain and the range of the function.
(lSltiG THE MfiltilDEfiS
Example 1 Solve -Jx + 3
-Jx + 3
(-Jx + 3)2
x + 3
o
=
=
=
=
3x - 1.
3x - 1
(3x - 1)2
9x2 -
~ Square both sides.
6x + 1
= 9x2 - 7x - 2
~ Write in standard form.
b2 - 4ac = (_7)2 - 4(9)(-2) = 121 = 112
Since b 2 - 4ac > 0, the equation has two different
real roots. Since b 2 - 4ac is a perfect square, the
equation can be solved by factoring.
9x2 - 7x - 2 = 0
(9x + 2)(x - 1) = 0
9x + 2 = 0 or x-I
Caution: If b2
.",~
-
4ac < 0, the
roots are complex conjugates; if
b2 - 4ac = 0, there is one real
double root.
= 0
x = -~or x = 1
9
-Jx + 3
Check:
=
3x - 1
-Jx + 3
+ 3 ~ 3( -~) - 1 -J 1 +
~-~
5
3:;t
-35
3
=
Z: 3(1) - 1
-{4 =
2.
3x - 1
2
.",~
Caution: Squaring both
sides of an equation does
not give an equation with
the same solutions as the
original equation. Always
check each possible root.
1 is a root.
-91S not a root.
The only solution is 1.
Example 2 Sketch the graphs of the line 4x - y = 9 and the parabola
y
2(x - 1) 2 - 3. Find the coordinates of any points of
intersection. Use algebra to justify your answer.
=
mm:mm • Graph the line using the intercepts and one other point.
• To graph the parabola, begin by identifying the vertex:
y = 2(x - 1)2 - 3 ~ y = 2(x - 1)2 + (-3)
y = a(x - h)2 + k
Thus, h
8
=
1 and k
ADVANCED MATHEMATICS
= - 3.
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The vertex is (1, -3) and the axis of symmetry is x = 1. Since the
value of a is 2, which is positive, the parabola is U-shaped.
Calculate a few points on one side of the y-axis and use symmetry to
plot the corresponding points on the other side of the y-axis. Sketch
the parabola by connecting the points. The sketch does not allow
you to tell for certain if there are 0, 1, or 2 solutions.
• 4x -
=
y
9
---7
=
Y
4x - 9
~ Solve the linear equation for
-2
y.
y = 2(x - 1)2 - 3
4x - 9 = 2(x2 - 2x + 1) - 3 ~ Substitute for y.
= 2x2 - 8x. + 8
o = 2(x2 - 4x + 4)
o
o =
o =
y
=
=
x - 2
=
x
4x -
9
2
2)2
(x -
4(2) - 9
=
~ Divide both sides by 2; factor.
~ Take the square root of each side.
-1
There is one point of intersection, (2, -1).
Exercises
1 - 2i
6 . S·Imp lif
I y3=j.
7. Solve..J6 - 2x
=
x - 3.
8. Sketch the graph ofy = (x + 1)2 + 3. Label the vertex, axis of
symmetry, and intercepts of the parabola.
9. Writing Explain how the discriminant can be used to show that the
line 4x + y = 4 and the parabola y = 2x2 - 5x + 6 do not intersect.
10. Application The manager of a new restaurant needs to set the price for
the daily dinner special. She has estimated the revenue, R, that the
restaurant could expect for each of three different prices, p.
p
R
$11
$2640
$14
$2856
$18
$2808
a. Explain why the data is or is not
reasonable to you.
b. Would a linear or quadratic model
fit the data values better? Why?
c. Find a function that models how
the revenue, R, varies according to
the price, p, charged.
d. What price will maximize the
revenue?
-f:z
11. Is the statement
_fa
{b = --V b
always,
sometimes, or never true when a and
b are real numbers and b '" O? Explain.
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ADVANCED MATHEMATICS
Student Resource Guide