Download Chapter 5: Quadratic Equations and Functions

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Chapter 5: Quadratic Equations and Functions
5.1 Modeling Data With Quadratic Functions
Quadratic Functions and Their Graphs
Note: If a = 0, it would be a
linear equation, f ( x) = bx + c .
Definition: Standard Form of a Quadratic Function
The equation f ( x) = ax + bx + c is a quadratic function if a ≠ 0 .
2
ax 2 is called the quadratic term
bx is called the linear term
c is called the constant term
The domain of a quadratic
function is all real numbers.
Example 1 Classifying Functions
Determine whether each function is linear or quadratic. Identify the quadratic, linear, and constant terms.
b. f ( x) = 3( x − 2 x) − 3( x − 2)
2
a. f ( x) = (2 x + 3)( x − 4)
= x(2 x + 3) − 4(2 x + 3) distribute
= 2 x + 3x − 8 x − 12
2
f ( x) = 2 x − 5 x − 12
2
distribute again
combine like terms
Is quadratic, where a = 2,
b = –5 & c = –12
Quadratic term: 2x 2
Linear term: −5x
Constant term: −12
2
3x 2 − 6 x − 3x 2 + 6 distribute
f ( x) = −6 x + 6 combine like terms
=
a = 0 so it’s not quadratic!
Look at the form, y = mx + b . The function is
actually linear with a slope of –6 and y-intercept
of 6.
Vocabulary for Quadratic Functions:
The graph of a quadratic function is a parabola.
The axis of symmetry is the line that divides a parabola into two parts that are mirror images. It is always a
vertical line defined by the x-coordinate of the vertex.
Points on the parabola have corresponding points on its mirror image. The two corresponding points are the
same distance from the axis of symmetry.
The vertex of a parabola is the point at which the parabola intersects the axis of symmetry. Also, it is where the
curve turns from decreasing (downhill) to increasing (uphill) or visa versa.
The y-value of the vertex of a parabola represents the maximum or minimum value of the function.
Example 2 Points on a Parabola
Use the graph of f ( x) = 2 x − 8 x + 8 . Identify the vertex, axis of symmetry,
2
points P′ and Q ′ corresponding to P and Q, and the range of f ( x) .
Vertex: ( 2, 0 )
Axis of Symmetry: x = 2 (a vertical line)
P (1, 2 ) → P′ ( 3, 2 ) both are 1 unit from x = 2 .
Q ( 0,8) → Q′ ( 4,8) both are 2 units from x = 2 .
Range: All real numbers where y ≥ 0
S. Stirling
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Using Quadratic Models
Example 3 Fitting a Quadratic Function to 3 Points
Find a quadratic function that includes the values in the table.
Note the ( x, y ) in the quadratic form y = ax + bx + c represent the points on the parabola.
2
Point
( 2,3)
( 3,13)
( 4, 29 )
Substitute in for x & y
Simplify
3 = a(2) + b(2) + c
→
3 = 4a + 2b + c
13 = a(3)2 + b(3) + c
→
13 = 9a + 3b + c
29 = a(4)2 + b(4) + c
→
29 = 16a + 4b + c
2
Now you need to solve the system of
equations for a, b & c. You do this in
the same way you solved systems
before, you simply need to substitute
or eliminate variables for 3 equations
and 3 variables.
Easiest variable to eliminate is c. Just multiply both sides of E2, “equation 2”, by –1. Then add E1 to E2 then
E2 to E3.
Solve:
Add E1 & –1*E2
E1 3 = 4a + 2b + c
E2 13 = 9a + 3b + c
E3 29 = 16a + 4b + c
3 = 4a + 2b + c
−13 = −9a − 3b − c
−10 = −5a − 1b
Take the Results:
Add E3 & –1*E2
Now can you find
a & b?
29 = 16a + 4b + c
−13 = −9a − 3b − c
16 = 7a + 1b
Check on the calculator:
Plot the 3 points. [STAT]
Put the equation in [Y=]
Were all three points on your graph?
Try [ZOOM] 9 or [ZOOM] out!
Calculator:
If you want the calculator to fit
your points…
−10 = −5a − 1b
16 = 7a + 1b
−10 = −5a − 1b
16 = 7a + 1b
6 = 2a
a=3
Find b:
16 = 7(3) + 1b
16 = 21 + b
b = −5
Find c:
3 = 4(3) + 2(−5) + c
The quadratic function is:
y = ax 2 + bx + c
y = 3x − 5 x + 1
2
3 = 12 − 10 + c
c =1
Example 4 Application
The table shows the height of a column of water as it drains from its
container. Model the data with a quadratic function. Graph the data and the
function. Use the model to estimate the water level at 35 seconds.
[STAT] EDIT
[STAT PLOT]
[STAT] CALC
5: QuadReg
It gives you the a, b & c!!
Put in [Y=]
and graph!
y = 0.0092 x 2 − 2.1036 x + 120.3333
Try the Quick Check page 240
answer: y = −2 x + 3x − 1 .
2
S. Stirling
[TABLE] find when X = 35
Or [CALC] 1: value type X = 35
Gives Y1 = 57.937 mm
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5.2 Properties of Parabolas
Properties of Parabolas
2
The graph of f ( x) = ax + bx + c is a parabola when a ≠ 0 .
When a > 0 , the parabola opens up.
When a < 0 , the parabola opens down.
The axis of symmetry is the line x = −
The x-coordinate of the vertex is −
b
.
2a
b
.
2a
The y-coordinate of the vertex is the y value of the function
b
or y =
2a
The y-intercept is ( 0, c ) .
when x = −
⎛ b ⎞
f ⎜− ⎟.
⎝ 2a ⎠
Example 1 Alt. Graphing Quadratics
y
Graph and label y = f ( x) = −4 x + 9
2
10
8
Identify a = −4 , b = 0 and c = 9 .
6
a = −4 , so a < 0 , the parabola opens down
Axis of symmetry is x = −
4
0
b
=−
=0.
2a
2(−4)
So the x-coordinate of the vertex is 0
2
and the y-coordinate of the vertex is f (0) = −4(0) + 9 = 9
The vertex is at ( 0,9 ) .
The y-intercept is ( 0, c ) , so ( 0,9 ) .
2
−2
−1
−2
x
1
2
3
−4
−6
8
More points on the graph.
x
y
2
1
−4(1) + 9 = 5
–1
−4(−1)2 + 9 = 5
2
−4(2)2 + 9 = −7
–2
−4(−2)2 + 9 = −7
Notice the symmetry!!
2
Try: y = 2 x − 4
S. Stirling
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Example 2 Graphing Quadratics
y
Graph and label y = x − 2 x − 3
2
12
Identify a = 1 , b = −2 and c = −3 .
10
8
a = 1 , so a > 0 , the parabola opens up
6
b
−2
=−
=1.
Axis of symmetry is x = −
2a
2(1)
4
2
So the x-coordinate of the vertex is 1
2
and the y-coordinate of the vertex is f (1) = (1) − 2(1) − 3 = −4
The vertex is (1, −4 ) ,
x
−3 −2 −1
−2
2
The y-intercept is ( 0, c ) , so f (0) = (0) − 2(0) − 3 = −3 , ( 0, −3) .
1
2 3 4 5 6
−4
6
More points on the graph.
x
y
1
−4
2
(2)2 − 2(2) − 3 = −3
Find points from
the symmetry.
x
y
0
−3
3
(3)2 − 2(3) − 3 = 0
–1
0
4
(4) − 2(4) − 3 = 5
–2
5
Try:
2
y = − x2 + 4x + 2
Finding Maximum and Minimum Values
Example 3 Finding a Minimum Value
What is the minimum value of the function?
f ( x) = 3x 2 + 12 x + 8
y
8
Identify a = 3 , b = 12 and c = 8 .
6
a = 3 , so a > 0 , the parabola opens up
4
2
(12)
= −2 .
Axis & the x-coordinate of vertex is x = −
2(3)
2
The y-coordinate is f (−2) = 3(−2) + 12(−2) + 8 = −4
So the minimum value is –4, when x = –2.
S. Stirling
−8
−6
−4
−2
x
2
4
−2
−4
6
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Example 4 Application
A company knows that −2.5 p + 500 models the number it sells per month of a certain make of unicycle,
where the price p can be set as low as $70 or as high as $120. Revenue from sales is the product of the price
and the number sold. What price will maximize the revenue? What is the maximum revenue?
Independent variable: p = price, Dependent variable: R(p) = revenue
From problem: Revenue = price * number sold
Substitute:
R( p) = p ( −2.5 p + 500 )
R( p) = −2.5 p 2 + 500 p
(500)
p=−
= 100 ,
Find the vertex:
2(−2.5)
R(100) = 100 ( −2.5 •100 + 500 ) = 100 • 250 = 25,000
Standard Form:
y
20000
10000
x
−50
50
100
150
200
25
A price of $100 will maximize the revenue of $25,000.
Example 4 Application Part 2
The number of widgets the Woodget Company sells can be modeled by −5 p + 100 , where p is the price of a
widget. What price will maximize revenue? What is the maximum revenue?
Independent variable: p = price per widget, Dependent variable: R(p) = revenue
Now Know:
Substitute:
Revenue = price * number sold
R( p) = p ( −5 p + 100 )
R( p) = −5 p 2 + 100 p
(100)
p=−
= 10 ,
Find the vertex:
2(−5)
R(10) = 10 ( −5 •10 + 100 ) = 10 ( 50 ) = 500
Standard Form:
A price of $10 will maximize the revenue of $500.
S. Stirling
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5.3 Transforming Parabolas
In Chapter 2, you learned to graph absolute value functions as transformations of their parent function,
y = x . Similarly, you can graph a quadratic function as a transformation of the parent function y = x 2 .
Summary Translations “Shifts”
Summary
Parent Function:
y=x
2
Vertical Translations
Translate up k units (k positive):
y = ax 2
Vertical Stretch a > 1
Stretch away from x-axis by a factor of a.
Vertical Shrink (fraction of) 0 < a < 1
Stretch away from x-axis by a factor of a.
Reflection in x-axis (negative) a < 0
Reflects over the x-axis and stretches or
shrinks.
Translate down k units (k positive):
Horizontal Translations (counter intuitive)
Translate right h units (h positive): y = ( x − h )
2
y = ( x + h)
2
Translate left h units (h positive):
1
y = − x2
2
2
Parent Function: y1 = x , with vertical shrink (factor ½ ) and reflect
y5 = ( x + 3)
x
–3
–2
–1
0
1
2
3
Shrink & Reflect: y3 = −
Parent
Shrink
Flip
y1
9
4
1
0
1
4
9
y2
4½
2
½
0
½
2
4½
y3
–4 ½
–2
–½
0
–½
–2
–4 ½
Example C: Graph
S. Stirling
−2
4
4
2
x
2
−2
−2
6
x
2
−2
4
6
−4
−6
−8
−6
10
−8
10
8
Example D: Graph
y
6
4
2
−2
4
−4
−4
2
−4
y
6
6
2
−4
and
.
8
y
y6 = 2 ( x + 3) − 6
The blue is the
stretch by
a factor of 2.
The pink shifts the
blue graph left 3
and down 6.
1
y1
2
8
2
For y4, shift parent up 4.
For y5, shift parent to the left 3.
over x-axis.
1
y1
2
y4 = x 2 + 4
Example B: Graph
Example A: Graph
Shrink: y2 =
y = x2 + k
y = x2 − k
−2
x
2
4
6
y7 = − ( x − 2 ) + 5
2
The blue is the
reflected over the
x-axis.
8
The pink shifts the
blue graph right 2
and up 5.
2
y
6
4
−4
−2
−2
−4
−4
−6
−6
−8
−8
10
10
x
2
4
6
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To transform the graph of a quadratic function, you can use the vertex form of a quadratic
function, y = a ( x − h ) + k .
2
Vertex Form of a Quadratic Function
The equation y = a ( x − h ) + k is a quadratic function if a ≠ 0 .
2
The graph (and vertex) of y = ax shifts h units horizontally and k units vertically.
2
The vertex is ( h, k ) and the axis of symmetry is the line x = h .
Example 1 (Method 1):
1
2
Graph y = − ( x − 2 ) + 3
2
1 2
Graph y1 = − x blue
2
Example 1 (Method 2): Graph
Shift “obvious” points to the right 2
and up 3, pink
Vertex ( h, k ) is ( 2,3)
8
Use the form to…
y
1
2
( x − 2) + 3
2
2
y = a ( x − h) + k
y=−
Graph the vertex and axis of symmetry.
Axis is x = 2 .
8
Find points and graph
them and their reflections.
6
6
4
4
2
−4
−2
x
0
x
2
−2
4
6
2
y
-2
−4
−6
y
−
1
2
( 0 − 2) + 3 = 1
2
−
1
2
( −2 − 2 ) + 3 = −5
2
−2
−2
x
2
4
6
8
−4
−6
−8
−8
10
10
Example 2: Writing an Equation for a Parabola
Write the equation for the parabola.
Use the vertex form: y = a ( x − h ) + k .
2
The vertex is ( 3, 4 ) or ( h, k ) → y = a ( x − 3) + 4
2
Your missing a, but you know a point on the parabola ( 5, −4 ) is ( x, y )
Substitute and find a.
(−4) = a ( (5) − 3) + 4
2
Use plot pattern too!
−4 = a ( 2 ) + 4
−8 = 4a
−2 = a
2
The equation is y = −2 ( x − 3) + 4
2
S. Stirling
Over
a=1
a = –2
0
±1
±2
±3
up 0
down 0
up 1
down 2
up 4
down 8
up 9
down 18
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Connections:
Both the vertex form and the standard form
give useful information about a parabola.
The standard form makes it easy to identify
the y-intercept.
The vertex form makes it easy to identify
the vertex and the axis of symmetry, and to
graph the parabola as a transformation of the
parent function. The graph shows the
relationship between the two forms.
Example 4: Writing an Equation for a Parabola
Write y = 2 x 2 + 10 x + 7 in vertex form.
Find a: a = 2 from the standard form.
Find the x-coordinate of the vertex: x = −
b
10
5
=−
= − = −2.5
2a
2(2)
2
2
⎛ 5⎞
⎛ 5⎞
⎛ 5⎞
Find the y-coordinate of the vertex: f ⎜ − ⎟ = 2 ⎜ − ⎟ + 10 ⎜ − ⎟ + 7
⎝ 2⎠
⎝ 2⎠
⎝ 2⎠
25
= 2i − 25 + 7 = 12.5 − 25 + 7 = −5.5
4
The vertex is ( −2.5, −5.5) or ( h, k ) .
Now just substitute into vertex form: y = a ( x − h ) + k
2
y = 2 ( x − (−2.5) ) + (−5.5)
2
Then simplify y = 2 ( x + 2.5) − 5.5
2
2
Try: Write y = −3x + 12 x + 5 in vertex form.
S. Stirling
Answer: y = −3 ( x − 2 ) + 17
2
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Example 3: Civil Engineering
The photo shows the Verrazano–Narrows Bridge in New York, which has the longest span of any suspension
bridge in the United States. A suspension cable of the bridge forms a curve that resembles a parabola. The curve
can be modeled with the function f ( x) = 0.0001432( x − 2130)2 , where x and y are measured in feet. The origin of
the function’s graph is at the base of one of the two towers that support the cable. How far apart are the towers?
How high are they?
Start by drawing a diagram.
The function is in vertex form. Since h = 2130
and k = 0, the vertex is at (2130, 0). The vertex
is halfway between the towers, so the distance
between the towers is 2(2130 ft) = 4260 ft.
To find the tower’s height, find y for x = 0 .
f (0) = 0.0001432(0 − 2130)2 ≈ 650
The towers are 4260 ft apart and about 650 ft high.
Example 3 Part 2: Civil Engineering
Suppose the towers in Example 3 are 4000 ft apart and 600 ft high. Write a function that could model the curve of
the suspension cable.
Use vertex form. Since the distance between the towers is 4000. The vertex is at (2000, 0).
now y = a( x − 2000) 2 + 0 or y = a( x − 2000) 2
Need to find a, and you know a point on the graph (0, 600).
substitute 600 = a(0 − 2000)2 and solve for a.
600 = 4000000a
a = 0.00015
The equation is y = 0.00015( x − 2000) 2
S. Stirling
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5.4 Factoring Quadratic Expressions
Finding Common and Binomial Factors
Factoring is rewriting an expression as the product of its factors. The greatest common factor (GCF) of
an expression is a common factor of the terms of the expression. It’s the common factor with the greatest
coefficient and the greatest exponent. You can factor any expression that has a GCF not equal to 1.
Example 1 Plus: Finding Common Factors
Completely factor each expression.
Strategy: Find the greatest common factor and “un-distribute”!
a. 4 x + 20 x − 12
b. 9n − 24n
2
2
(
2
= 4 x + 5x − 3
)
= 3n ( 3n − 8)
A quadratic trinomial is an expression in the form ax 2 + bx + c . You can factor many quadratic trinomials
into two binomial factors.
Example 2: Factoring
Factor x 2 + 8 x + 7
Step 1: Find two factors with a product of ac and a sum b.
Step 2: Rewrite the bx term with the factors you found.
find
guess
product ac
1• 7 = 7
sum b
8
1• 7 = 7
1+ 7 = 8
x2 + 8x + 7
x 2 + 1x + 7 x + 7
Step 3: Group and factor out the GCF from each grouping
(x
2
+ 1x ) + ( 7 x + 7 )
x ( x + 1) + 7 ( x + 1)
Step 4: Factor out the common
binomial factor
4 terms
x ( x + 1) + 7 ( x + 1)
only 2 terms
( x + 7 )( x + 1)
Notice the binomial factors (x + 1) in both terms
Example 2 Quick Check:
Example 2 Quick Check:
a. Factor x + 6 x + 8
b. Factor x 2 + 12 x + 32
2
ac
b
find
guess
1• 8 = 8
2•4 = 8
2+4 = 6
6
ac
b
find
guess
1 • 32 = 32
4 • 8 = 32
4 + 8 = 12
12
x2 + 2x + 4x + 8
x ( x + 2) + 4 ( x + 2)
x 2 + 4 x + 8 x + 32
x ( x + 4) + 8 ( x + 4)
( x + 4)( x + 2)
( x + 8)( x + 4)
S. Stirling
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Example 3: Factoring (The factors do not need to be positive.)
Factor x 2 − 17 x + 72
Step 1: Find two factors with a product of ac and a sum b.
ac
b
find
guess
1 • 72 = 72
−9 • −8 = 72
−9 + −8 = −17
–17
Use prime factors:
72
2 • 36
2•6•6
2•2•3•2•3
Step 2: Rewrite the bx term as a sum with the factors you found.
x 2 − 17 x + 72
x 2 − 9 x + −8 x + 72
Step 4: Factor out the common
binomial factor
Step 3: Group and factor out the GCF from each grouping
(x
2
− 9 x ) + ( −8 x + 72 )
x ( x − 9 ) + −8 ( x − 9 )
( x − 8)( x − 9)
x ( x − 9 ) + −8 ( x − 9 )
Example 3 Quick Check:
Example 3 Quick Check:
a. Factor x − 6 x + 8
c. Factor x 2 − 11x + 24
2
ac
b
find
guess
1• 8 = 8
−2 • −4 = 8
−2 + −4 = −6
–6
ac
b
find
guess
1 • 24 = 24
−3 • −8 = 24
−3 + −8 = −11
–11
x 2 − 2 x + −4 x + 8
x ( x − 2 ) + −4 ( x − 2 )
x 2 − 3x + −8 x + 32
x ( x − 3) + −8 ( x − 3)
( x − 4)( x − 2)
( x − 8)( x − 3)
Example 4: More Factoring
(Same thing different numbers.)
Factor x − x − 12
ac
b
Step 2:
find
guess
1 • −12 = −12
3 • −4 = −12
3 + −4 = −1
–1
x 2 − x − 12
x 2 + 3x + −4 x − 12
Step 3: x ( x + 3) + −4 ( x + 3)
Step 4:
S. Stirling
24
2 • 12
2•2•2•3
Example 4 Quick Check:
a. Factor x 2 − 14 x − 32
2
Step 1:
Use prime factors:
ac
b
find
guess
1 • −32 = −32
2 • −16 = −32
2 + −16 = −14
–14
x 2 + 2 x + −16 x − 32
x ( x + 2 ) + −16 ( x + 2 )
( x − 16)( x + 2)
Use prime factors:
32
8•4
2•2•2•2•2
( x − 4)( x + 3)
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Example 4 Quick Check:
Example 4 Quick Check:
b. Factor x + 3x − 10
c. Factor x 2 + 4 x − 5
2
ac
b
find
guess
1 • −10 = −10
−2 • 5 = −10
−2 + 5 = 3
3
ac
b
find
guess
1 • −5 = −5
−1 • 5 = −5
−1 + 5 = 4
4
x 2 − 2 x + 5 x − 10
x ( x − 2) + 5 ( x − 2)
x 2 − 1x + 5 x − 5
x ( x − 1) + 5 ( x − 1)
( x + 5)( x − 2)
( x + 5)( x − 1)
Example 5: Factoring (Same stuff, but a bit harder.)
Factor 3x 2 − 16 x + 5
Step 1: Find two factors with a product of ac and a sum b.
ac
b
find
guess
3 • 5 = 15
−1 • −15 = 15
−1 + −15 = −16
–16
Step 2: Rewrite the bx term as a sum with the factors you found.
3x 2 − 16 x + 5
3x 2 − 15 x + −1x + 5
Step 3: Group and factor out the GCF from each grouping
( 3x
2
Step 4: Factor out the common
binomial factor
3x ( x − 5) + −1( x − 5)
− 15 x ) + ( −1x + 5)
( 3x − 1)( x − 5)
3x ( x − 5) + −1( x − 5)
Example 5 Quick Check:
Example 5 Quick Check:
a. Factor 2 x + 11x + 12
c. Factor 2 x 2 − 7 x + 6
2
ac
b
find
guess
2 •12 = 24
8 • 3 = 24
8 + 3 = 11
11
2 x 2 + 3x + 8 x + 12
x ( 2 x + 3 ) + 4 ( 2 x + 3)
( x + 4)( 2 x + 3)
S. Stirling
Use prime factors:
24
2 • 12
2•2•2•3
ac
b
find
guess
2 • 6 = 12
−4 • −3 = 12
−4 + −3 = −7
–7
2 x 2 − 4 x + −3x + 6
2 x ( x − 2 ) + −3 ( x − 2 )
( 2 x − 3)( x − 2)
Use prime factors:
12
2•6
2•2•3
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Example 6: Factoring (Same stuff, but a bit harder.)
Factor 4 x 2 − 4 x − 15
Use prime factors:
Step 1: Find two factors with a product of ac and a sum b.
ac
b
find
guess
4 • −15 = −60
−10 • 6 = −60
−10 + 6 = −4
–4
60
6 • 10
2•3•2•5
Step 2: Rewrite the bx term as a sum with the factors you found.
4 x 2 − 4 x − 15
4 x 2 − 10 x + 6 x − 15
Step 4: Factor out the common
binomial factor
Step 3: Group and factor out the GCF from each grouping
( 4x
2
− 10 x ) + ( 6 x − 15)
2 x ( 2 x − 5) + 3 ( 2 x − 5)
( 2 x + 3)( 2 x − 5)
2 x ( 2 x − 5) + 3 ( 2 x − 5)
Example 6 Quick Check:
Example 6 Quick Check:
a. Factor 2 x + 7 x − 9
b. Factor 3x 2 − 16 x − 12
2
ac
b
find
guess
2 • −9 = −18
−2 • 9 = −18
−2 + 9 = 7
7
ac
b
find
guess
3 • −12 = −36
−18 • 2 = −36
−18 + 2 = −16
–16
2x2 − 2x + 9x − 9
2 x ( x − 1) + 9 ( x − 1)
3x 2 − 18 x + 2 x − 12
3x ( x − 6 ) + 2 ( x − 6 )
( 2 x + 9)( x − 1)
( 3x + 2)( x − 6)
Use prime factors:
36
6•6
2•3•2•3
A perfect square trinomial is the product you obtain when you square a
Perfect Square Trinomials
binomial. An example is x 2 + 10 x + 25 , which can be written as ( x + 5) . The
2
( a + b ) = a 2 + 2ab + b2
2
( a − b ) = a 2 − 2ab + b2
2
first term and the third term of the trinomial are always positive, as they
represent the squares of the two terms of the binomial. The middle term of the
trinomial is two times the product of the terms of the binomial.
Pre-Example 7a:
a. Multiply ( x + 3)
means
Pre-Example 7b:
2
( x + 3)( x + 3)
x 2 + 3x + 3x + 9
x2 + 6x + 9
2
2
formula a + 2ab + b with a = x and b = 3
2
2
( x ) + 2 ( x )( 3) + ( 3) = x 2 + 6 x + 9
S. Stirling
b. Multiply ( 2 x − 5)
means
2
( 2 x − 5)( 2 x − 5)
4 x 2 − 10 x − 10 x + 25
4 x 2 − 20 x + 25
2
2
formula a − 2ab + b with a = 2x and b = 5
2
2
( 2 x ) − 2 ( 2 x )( 5) + ( 5) = 2 x 2 − 20 x + 25
Page 13 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
Now try expanding, multiplying, the perfect square trinomials simply by using the formulas.
Pre-Example 7c:
Pre-Example 7d:
c. Multiply ( 3x + 4 )
d. Multiply ( x − 7 )
2
2
2
formula a + 2ab + b with a = 3x and b = 4
( 3x )
2
+ 2 ( 3x )( 4 ) + ( 4 )
2
2
2
2
formula a − 2ab + b with a = 1x and b = 7
( x)
9 x 2 + 24 x + 16
2
− 2 ( x )( 7 ) + ( 7 )
2
x 2 − 14 x + 49
To factor, using these formulas, all you need to do is identify a and b utilizing the formulas, then work it
in reverse.
Example 7:
Example 7 Quick Check:
Factor 9 x − 42 x + 49
a. Factor
2
since ( 3x ) = 9 x , a = 3x
4 x 2 + 12 x + 9
2
2
since ( 2 x ) = 4 x , a = 2 x
since ( 7 ) = 49 , b = 7
since ( 3) = 9 , b = 3
2
2
check a − 2ab + b
2
2
check a + 2ab + b
2
2
2
( 3x )
2
− 2 ( 3x )( 7 ) + ( 7 ) = 9 x 2 − 42 x + 49
2
2
factored is ( a − b ) = ( 3x − 7 )
2
2
( 2x )
2
+ 2 ( 2 x )( 3) + ( 3) = 4 x 2 + 12 x + 9
2
factored is ( a + b ) = ( 2 x + 3)
2
Example 7 Quick Check:
Example 7 Quick Check:
64 x − 16 x + 1
2
2
since ( 8 x ) = 64 x , a = 8 x
c. Factor
since (1) = 1 , b = 7
since ( 9 ) = 81 , b = 9
2
2
check a − 2ab + b
2
2
check a + 2ab + b
25 x 2 + 90 x + 81
2
2
since ( 5 x ) = 25 x , a = 5 x
2
b. Factor
2
(8x )
2
2
2
− 2 ( 8 x )(1) + (1) = 64 x 2 − 16 x + 1
2
factored is ( a − b ) = ( 8 x − 1)
2
2
( 5x )
2
+ 2 ( 5 x )( 9 ) + ( 9 ) = 25 x 2 + 90 x + 81
2
factored is ( a + b ) = ( 5 x + 9 )
2
2
Difference of Two Squares
( a + b )( a − b ) = a2 − b2
An expression of the form a 2 − b2 is defined as the difference of two
squares. It also follows a pattern that makes it easy to factor.
Pre-Example 8a:
Pre-Example 8b:
long way x − 3x + 3x − 9
long way 4 x − 10 x + 10 x − 25
x2 − 9
formula ( a + b )( a − b ) with a = x and b = 3
4 x 2 − 25
formula ( a + b )( a − b ) with a = 2x and b = 5
a. Multiply ( x + 3)( x − 3)
2
( x ) − ( 3)
2
S. Stirling
2
= x −9
2
b. Multiply ( 2 x + 5)( 2 x − 5)
2
( 2 x ) − ( 5)
2
2
= 4 x − 25
2
Page 14 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
Now use the formula in reverse.
Example 8 Quick Check:
Example 8 Quick Check:
x − 64
2
a = x and since ( 8) = 64 , b = 8
b. Factor
4 x 2 − 49
2
2
since ( 2 x ) = 4 x , a = 2 x
2
a. Factor
2
2
check a − b = ( x ) − ( 8) = x − 64
since ( 7 ) = 49 , b = 7
factored is ( a + b )( a − b ) = ( x + 8)( x − 8)
2
2
check a − b = ( 2 x ) − ( 7 ) = 4 x − 49
2
2
2
2
2
2
2
factored is ( a + b )( a − b ) = ( 2 x + 7 )( 2 x − 7 )
Example 8:
The photo shows the thin ring that is the cross-section of the pipe.
Find an expression, in factored form, that gives the area of the cross-section in completely
factored form.
Area of a washer = π R 2 − π r 2
Outer radius R = 3
Inner radius r
A = π ( 3) − π r 2 = 9π − π r 2
2
Factor out GCF: π ( 9 − r 2 )
Diff. 2 Squares: π ( 3 + r )( 3 − r )
Factor out a GCF.
Combined factoring:
1. Factor out a GCF. Make
highest power term positive.
2. Test for special case:
Perfect Square Trinomial:
a + 2ab + b = ( a + b )
2
2
a 2 − 2ab + b2 = ( a − b )
2
2
Difference of two squares:
a 2 − b2 = ( a + b )( a − b )
3. Factor by ac and b method.
−15 x 2 − 5 x
−5 x ( 3x + 2 )
A special case. Look for perfect squares.
4 x 2 − 28 x + 49 Is a = 2 x , b = 7 ?
2
2
2
Check a − 2ab + b = ( a − b )
( 2x)
2
− 2 ( 2 x )( 7 ) + ( 7 )
2
Factor ( a − b ) → ( 2 x − 7 )
2
Yes it checks!
2
2
Factor by ac and b method. y = ax + bx + c
10 x 2 − 17 x + 3
(10 x2 −15x ) + ( −2 x + 3)
5 x ( 2 x − 3) + −1( 2 x − 3)
Set up:
Find #s:
ac = 30 −15i−2
b = −17 −15 + −2
( 5x − 1)( 2 x − 3)
S. Stirling
Page 15 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
Algebra 1 Review: Square Roots and Radicals
A radical symbol
indicates a square root.
The expression 16 means the principal, or positive, square root of 16 or 4.
The expression − 16 means the negative square root of 16 or –4 .
In general,
x 2 = x or ± x for all real numbers x.
When simplifying radicals, three rules apply
1) You may not have factors that are perfect squares under the
radical sign.
The cure? Factor the radicand and use the multiplication
property.
Simplify
12 =
4•3 = 4 • 3 = 2 3
75 =
25 • 3 = 25 • 3 = 5 3
Square Root Properties
Multiplication Property of Square Roots
For any numbers a ≥ 0 and b ≥ 0 ,
ab = a • b
Division Property of Square Roots
For any numbers a ≥ 0 and b > 0 ,
a
a
=
b
b
72 = 9 • 8 = 9 • 4 • 2 = 9 • 4 • 2
= 3• 2 • 2 = 6 2
2) You may not have fractions under the radical sign.
The cure? Use the division property and simplify further, if necessary.
Simplify
4
=
9
4 2
=
9 3
50
=
49
50
25i2 5 2
=
=
7
7
49
3) You may not have radicals in the denominator of a fraction.
The cure? Multiply the numerator and denominator of the fraction by the radical in the denominator (to create a
perfect square). Simplify further if needed.
Simplify
3
3
2 3 2 3 2
•
=
=
=
2
2
2
4
2
−
2 5 2 5
3 2 15 2 15
•
=
=
=
3
3
3
3
9
5
7
35
35
5
•
=−
=−
= −
7
7
7
7
49
S. Stirling
Page 16 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
5.5 Solving Quadratic Equations
Solving an equation means to find the values that make the original sentence true.
With linear equations you usually only have one solution, with quadratics you usually have two.
Zero Product Property
If ab = 0 , then either a = 0 or b = 0 or both = 0.
You will use this property to help you solve
quadratics.
Example 1: Solving by Factoring
Example 1 plus: Solving by Factoring
Solve 2 x − 11x = −15
Solve 16 x = 8 x
2 x 2 − 11x = −15
2 x 2 − 11x + 15 = 0
get equation = 0
2
( 2 x − 6 x ) + ( −5x + 15) = 0 factor
16 x 2 = 8 x
16 x 2 − 8 x = 0 get equation = 0
8x ( 2 x − 1) = 0 factor & use Zero Product Prop.
8 x = 0 or 2 x − 1 = 0 so solve for xs
x = 0 or x = 1 2
2
2
2 x ( x − 3) + −5 ( x − 3) = 0
( 2 x − 5)( x − 3) = 0
now use Zero Product Prop.
2 x − 5 = 0 or x − 3 = 0
x = 5 or x = 3
2
so solve for x
Example 1 More Practice Solve by Factoring:
a. Solve
x + 7 x + 12 = 0
2
x 2 + 3x + 4 x + 12 = 0
x ( x + 3) + 4 ( x + 3) = 0
( x + 4)( x + 3) = 0
x + 4 = 0 or x + 3 = 0
x = −4 or x = −3
Caution!! Don’t try alternate procedures!!
16 x 2 = 8 x
16 x 2 8 x
=
divide both sides by 8x
8x
8x
2 x = 1 Oops! You lost a root! NO GOOD /
Example 1 More Practice Solve by Factoring:
b. Solve
x 2 + 7 x = 18
x 2 + 7 x − 18 = 0
x 2 + 9 x + −2 x − 18 = 0
x ( x + 9 ) + −2 ( x + 9 ) = 0
( x − 2)( x + 9) = 0
x − 2 = 0 or x + 9 = 0
x = 2 or x = −9
Example 1 More Practice Solve by Factoring:
c. Solve
−2 x 2 = 4 x − 6
0 = 2 x2 + 4 x − 6
0 = 2 ( x 2 + 2 x − 3)
factor out GCF and divide!
0 = x 2 + 3x + −1x − 3
0 = x ( x + 3) + −1( x + 3)
S. Stirling
0 = ( x − 1)( x + 3)
x − 1 = 0 or x + 3 = 0
x = 1 or x = −3
Page 17 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
Look at Example 2 below. How is it different from the previous two Examples? VIP! You guessed it!?
There is no linear term. So, you can solve by taking the square root.! Careful!! Don’t lose a root!
Definition:
Solve x = 25 .
Solve x = 25 .
By substitution, you know the
solution is +5 or –5, because
If you solve this by taking the square root
of both sides, you need to remember that
you get a positive and a negative root.
Write ± for “plus or minus”
2
52 = 25
and
2
( −5)
2
= 25
x 2 = 25
→
x 2 = x or ± x
for all real numbers x.
25 = 52 = 5 = 5 and
Note:
x 2 = ± 25 = ±5
25 =
( −5)
2
= −5 = 5
Example 2: Solving by Finding Square Roots
Example 2: Solving by Finding Square Roots
Hint: Work in the reverse order of operations
because you’re solving!
Solve 3x = 24
Solve 5 x − 180 = 0
2
5 x 2 − 180 = 0
Would be hard to factor, no?
2
5 x = 180
get the quadratic term alone!
2
5 x 180
2
=
isolate the x completely
5
5
2
x = 36
now take the square root, both sides
2
3x 2 = 24
3x 2 24
=
3
3
2
x =8
x=± 8
2
isolate the x completely
now take the square root, both sides
need the ±
x = ±2 2
simplify (take perfect squares out)
x 2 = ± 36 need the ± !!
x = ±6
simplify
Example 2 More Practice Solve by Square Root:
Alternate Method, Factoring:
a. Solve 4 x − 25 = 0
a. Solve 4 x − 25 = 0
4 x 2 − 25 = 0
4 x 2 = 25
4 x 2 25
=
4
4
25
x2 =
4
25
x2 = ±
4
5
x=±
2
4 x 2 − 25 = 0 Try factoring! Why not?
( 2 x + 5)( 2 x − 5) = 0 difference of two squares
2 x + 5 = 0 or 2 x − 5 = 0
2 x = −5 or 2 x = 5
5
5
x = − or x =
2
2
5
x=±
2
2
S. Stirling
2
Could factor, but…
get the quadratic term alone!
2
isolate the x completely
now take the square root, both sides
need the ± !!
simplify
See why you need to use the ± when using the square
root method? You would totally lose one of your
roots!
Page 18 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
Not every quadratic equation can be solved by factoring or by finding square roots. You can solve ax 2 + bx + c by
graphing y = ax 2 + bx + c , its related quadratic function. The value of y is 0 where the graph intersects the x-axis.
Each x-intercept is a zero of the function and a root of the equation.
Solving by Tables and/or by Graphing
By the way, you can use these methods to solve any types of equations! But, you need a calculator.
Enter the left hand part of the equation in Y1 and enter the right hand part of the equation in Y2, then
you’re just looking for when the y-values are the same! (The x-values that make the sentence true!)
Calculator Solutions [TABLE]:
[Y=] enter left hand part Y1
enter left hand part Y2
[TBLSET] start at 0 with ΔTbl = 1
[TABLE] Look for when Y1 = Y2.
You can tweek the ΔTbl to a smaller
number until the Y values get really close.
Calculator Solutions [GRAPH]:
[Y=] enter left hand part Y1
enter left hand part Y2
[GRAPH]
[ZOOM] choose a window if necessary.
Look for both intersections if possible
2nd [CALC] 5: intersect
Answer calculator’s questions.
Make sure you find both answers if there
are two points of intersections.
Example 4 alt: Solving by Tables
Solve 2 x − 11x = −15
2
Example 5 alt: Solving by Graphing
Solve x − 2 x = 4
2
Y1 = 2 x − 11x
Y2 = −15
2
[TBLSET] TblStart = 0 with ΔTbl = 1
Look for when Y1 = Y2.
Happens at x = 3, but you know there is probably
another solution, so change ΔTbl = 0.1.
Look again…
At x = 2.5, there is another solution.
With quadratics, there is a maximum of 2
solutions.
Y1 = x 2 − 2 x
Y2 = 4
[GRAPH] Try [ZOOM] 6:
Look for when Y1 = Y2, the intersection points.
2nd [CALC] 5: intersect Guess at the left intersection.
For the left intersection: X = –1.236 Y = 4
2nd [CALC] 5: intersect Guess at the right intersection.
For the right intersection: X = 3.236 Y = 4
With quadratics, there is a maximum of 2
solutions.
Example 4: Solving by Tables
Solve x − 5 x + 2 = 0
2
Y1 = x 2 − 5 x + 2
Y2 = 0
[TBLSET] TblStart = 0 with ΔTbl = 1
Look for when Y1 = Y2.
Notice that the sign changes between x = 0 and x =
1, then again between x = 4 and x = 5, so must be
near zero between those values.
So change ΔTbl = 0.01.
Look again near the x-values you found before…
At x ≈ 0.21 and x ≈ 4.56 are the solutions.
S. Stirling
Page 19 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
5.6 Complex Numbers
Introduction:
The imaginary number i is defined as the number whose
Solve x = −25 .
square is –1. So i = −1 and i =
Try to solve… x = ± −25 , but you
can’t take the square root of a
negative…well not if the answer needs
to be a real number.
An imaginary number is any number of the form a + bi,
where a and b are real numbers, and b ≠ 0.
So to simplify complex numbers, you will rewrite
any −a as i a , then simplify as you would do
normally.
Example:
−4 = −1 • 4 = i 4 = 2i
Note that
2
2
2
correct! So, you must write it as an imaginary
number first, before you simplify.
Example 1 QC:
a. Simplify
−7 .
Example 1 QC:
b. Simplify
Square Root of a Negative Real Number
−12 .
Simplify
−8
−1i4i2 Factor.
2i 2
Take out the factor of –1 and 4.
Example 1 QC:
c. Simplify
−8 by using the imaginary number i.
−36 .
Note:
Now that
a is defined for a < 0 ,
explain why
−1i7
i 7
−a = i a .
For any positive real number a,
Example 1: Simplifying Numbers Using i
( −4 ) = ( 2i ) = 4i = 4 • −1 = −4
NOT ( −4 ) ≠ −4 • −4 ≠ 16 = 4 not
2
−1 .
2
2
−1i4i3
2i 3
−1i36
6i
a2 = a
a2 ≠
and
( a) .
( a)
2
2
=a
Imaginary numbers and real numbers together make up the set of complex numbers.
Example 2: Simplifying Imaginary
Numbers
Write the complex number
the form a + bi.
Now we can expand our number types
−9 + 6 in
−9 + 6
3i + 6 Simplify the radical expression.
6 + 3i
Write in the form a ± bi.
Example 2 Quick Check:
Write the complex number
the form a + bi.
−18 + 7 in
−1i9i2 + 7
7 + 3i 2
S. Stirling
Page 20 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
2
2
OPTIONAL: The absolute value of a complex number is a + bi = a + b .
Example 3: Finding Absolute Value
a. Find 5i .
a. Find
6 − 4i = 62 + ( −4 )
5i = 02 + 52 = 5
6 − 4i .
2
36 + 16 = 52 = 4i13 = 2 13
b. Find 3 − 4i .
3 − 4i = 32 + ( −4 )
Example 3 QC:
Example 3 QC:
2
−2 + 5i =
b. Find
( −2)
2
−2 + 5i
.
+ 52
4 + 25 = 29
9 + 16 = 25 = 5
You can apply the operations of real numbers to complex numbers. Remember, if the sum of two complex
numbers is 0, then each number is the opposite, or additive inverse, of the other.
Example 4 QC:
a. Find the additive
inverse of −5i
Example 4: Additive Inverse of a
Complex Number
Find the additive inverse of −2 + 5i
− ( −5i )
5i
− ( −2 + 5i ) Find the opposite.
2 − 5i Simplify.
Example 4 QC:
b. Find the additive
inverse of 4 − 3i
− ( 4 − 3i )
−4 + 3i
Example 4 QC:
c. Find the additive
inverse of a + bi
− ( a + bi )
−a − bi
To add or subtract complex numbers, combine the real parts and the imaginary parts separately.
Example 5: Adding Complex
Numbers
Simplify the expression −2 + 5i
( 5 + 7i ) + ( −2 + 6i )
5 − 2 + 7i + 6i
3 + 13i
Example 5 QC:
Example 5 QC:
b. Simplify
a. Simplify
Example 5 QC:
c. Simplify
7 − ( 3 + 2i )
7 − 3 − 2i
4 − 2i
(8 + 3i ) − ( 2 + 4i )
8 + 3i − 2 − 4i
8 − 2 + 3i − 4i
6−i
( 4 − 6i ) + 3i
4 − 6i + 3i
4 − 3i
For two imaginary numbers bi and ci , ( bi )( ci ) = bci = bci−1 = −bc .
2
Example 6: Multiplying Complex
Numbers
a. Find ( 5i )( −4i )
−20i 2 and since i 2 = −1
−20i−1 = 20
b. Find ( 2 + 3i )( −3 + 5i )
−6 + 10i − 9i + 15i 2 and since i 2 = −1
−6 − 15 + 10i − 9i
−21 + i
S. Stirling
Example 6 QC:
a. Simplify
(12i )( 7i ) = 84i 2 = −84
Example 6 QC:
Example 6 QC:
b. Simplify
c. Simplify
24 − 18i − 20i + 15i 2
24 − 15 − 18i − 20i
9 − 38i
16 + 12i − 36i − 27i 2
16 + 27 + 12i − 36i
43 − 24i
( 6 − 5i )( 4 − 3i )
( 4 − 9i )( 4 + 3i )
Page 21 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
Some quadratic equations have solutions that are complex numbers.
Example 7: Finding Complex Solutions
Solve 4 x 2 + 100 = 0
4 x 2 −100
2
=
Isolate the x .
4
4
2
x = −25 Now take the square root of both sides.
x = ± −25 Remember the ± (two roots).
x = ±5i
x = 5i
Check:
Example 7 Quick Check:
a. Simplify
Solve 3x 2 + 48 = 0
3x 2 −48
=
3
3
2
x = −16
x = ± −16
x = ±4i
Show it works!
4 ( 5i ) + 100 = 0
2
4i25i 2 + 100 = 0
−100 + 100 = 0
Check:
x = −5i
4 ( −5i ) + 100 = 0
2
4i25i 2 + 100 = 0
−100 + 100 = 0
Example 7 Quick Check:
Show it works!
b. Simplify
Solve −5 x 2 − 150 = 0
−5 x 2 150
=
−5
−5
2
x = −30
x = ± −30
x = ±i 30
Example 7 Quick Check:
c. Simplify
Solve 8 x 2 + 2 = 0
8 x 2 −2
=
8
8
1
x2 = −
4
1
1
x = ±i
=± i
4
2
S. Stirling
Page 22 of 29
Ch 5 Alg 2 Note Sheet Key
Name _______________________
5.7 Completing the Square
Pre-Example: Perfect Squares
= x + 6x + 9
2
Given: x + 10 x
Geometrically:
Given: x + 12 x
Geometrically:
2
x2
3x
x
3x
9
5 +
x
+5
x2
5x
5x
?
x
x
+3
x
x
What would you need to
complete the square?
2
Geometrically:
3 +
What would you need to
complete the square?
6 +
( x + 3)
2
If you have x + 6 x , you need
+6
x2
6x
6x
?
2
the +9 to complete the square.
x 2 + 6 x + 9 = ( x + 3)
Add 36 to complete the square.
Add 25 to complete the square.
x 2 + 10 x + 25 = ( x + 5)
2
x 2 + 12 x + 36 = ( x + 6 )
2
2
The process of completing the square can be used to solve quadratic equations as well as used to rewrite
equations into vertex form. Look for patterns above. Take half of the coefficient of the x-term, then add
the square of it to create a trinomial.
Example 1: Solving a Perfect Square
Trinomial Equation
Solve x 2 + 10 x + 25 = 36
x + 10 x + 25 = 36
2
( x + 5)
2
= 36
Example 1: QC
Solve x 2 − 14 x + 49 = 81
x 2 − 14 x + 49 = 81
The left side is a perfect square.
Rewrite.
x + 5 = ±6
Now take the square root of both sides.
Solve the resulting equations:
x + 5 = 6 and x + 5 = −6
x = 1 and x = −11
( x − 7)
2
= 81
x − 7 = ±9
x − 7 = 9 and x − 7 = −9
x = 16 and x = −2
Take half of b.
Square it and
add it on.
⎛b⎞
x + bx + ⎜ ⎟
⎝2⎠
2
2
S. Stirling
Page 23 of 29