Chapt 8
Quadratic Equations & Functions
8.1 Completing a Square
• Given: x2 = u
x = +√(u) or
x = -√(u)
• E.g.
Given: x2 = 3
x = √(3) or
x = -√(3)
Completing the Square
• Solve: x2 – 6x + 4 = 0
• x2 – 6x = -4
• How to make the left side a perfect square?
x2 – 6x + 9 = -4 + 9
(x – 3)2 = 5
x – 3 = √5
or x – 3 = -√5
x = 3 + √5
or x = 3 - √5
• Check:
(3 + √5)2 – 6(3 + √5) + 4 = 0 ?
9 + 6√5 + 5 – 18 - 6√5 + 4 = 0 ?
(9 + 5 – 18 + 4) + (6√5 - 6√5 ) = 0 yes
Completing the Square
• Solve: 9x2 – 6x – 4 = 0
• (9x2 – 6x – 4)/9 = 0/9
x2 – (6/9)x – 4/9 = 0
x2 – (2/3)x = 4/9
x2 – (2/3)x + (1/9) = 4/9 + (1/9)
(x – 1/3)2 = 5/9
x – 1/3 = √(5/9)
or x – 1/3 = -√(5/9)
x = 1/3 + √(5)/√(9) or x = 1/3 - √(5)/√(9)
x = 1/3 + √(5)/3
or x = 1/3 - √(5)/3
x = (1 + √(5))/3
or x = (1 - √(5))/3
Your Turn
•
Solve by completing the square:
x2 + 3x – 1 = 0
•
x2 + 3x = 1
x2 + 3x + 9/4 = 1 + 9/4
(x + 3/2)2 = 13/4
x + 3/2 = ± √(13/4)
x = -3/2 ± √(13)/2
x = (-3/2 ± √(13)) / 2
Your Turn
•
•
Solve by completing the square:
3x2 + 6x + 1 = 0
3x2 + 6x = -1
(3x2 + 6x) / 3 = -1/3
x2 + 2x = -1/3
x2 + 2x + 1 = -1/3 + 1
(x + 1)2 = 2/3
x + 1 = ±√(2/3)
x = 1 ±√(2/3)
x = 1 ±√(2/3) √(3)/√(3)
x = 1 ±√(6) / 3
x = (3 ±√(6) ) / 3
Review
• Solve by completing the square.
(4x – 1)2 = 15
• 16x2 – 8x + 1 = 15
• 16x2 – 8x = 14
• (16x2 – 8x)/16 = 14/16
• x2 – (1/2)x = 7/8
• x2 – (1/2)x + 1/16 = 7/8 +1/16
• (x – ¼)2 = 15/16
• x – ¼ = ±√(15/16)
• x = ¼ + √(15/16) = ¼ + √(15)/4 = (1 + √(15)/4
• x = ¼ - √(15/16) = ¼ - √(15)/4 = (1 - √(15)/4
8.2 Quadratic Formula
• Given: ax2 + bx + c = 0, where a > 0.
• x2 + (b/a)x + (c/a) = 0
x2 + (b/a)x = -(c/a)
x2 + (b/a)x + b2/(4a2) = -(c/a) + b2/(4a2)
(x + (b/(2a))2 = -(c/a) + b2/(4a2)
(x + (b/(2a))2 = -(c/a)(4a2)/(4a2) + b2/(4a2)
(x + (b/(2a))2 = -((c/a)(4a2) + b2) / (4a2)
x + b/(2a) = ±√ ((b2 – 4ac)/(4a2))
x = (-b /(2a) ±√ ((b2 – 4ac)/(2a)
x = (-b ±√ (b2 – 4ac)) / (2a)
Quadratic Formula
• Given: ax2 + bx + c = 0, where a > 0.
•
-b ± √(b2 – 4ac)
x = ----------------------2a
• E.g., if
3x2 – 2x – 4 = 0
a = 3; b = -2; c = -4
•
2 ±√(4 + 48) 2 ± √(52) 1 ± √(13)
x = ---------------- = ------------ = --------------6
6
3
Application
• The number of fatal vehicle crashes has been
found to be a function of a driver’s age. Younger
and older driver’s tend to be involved in more
fatal accidents, while those in the 30’s and 40’s
tend to have the least such accidents.
• The number of fatal crashes per 100 million
miles, f(x), as a function of age, x, is given by
f(x) = 0.013x2 – 1.19x + 28.24
• What age groups are expected to be involved in
3 fatal crashes per 100 million miles driven?
Fatal Crashes vs Age of Drivers
Fatal Crashes / 100 Million
Miles of Driving
Age of US Drivers and Fatal Crashes
20.0
18.0
16.0
14.0
12.0
10.0
8.0
6.0
4.0
2.0
0.0
17.7
16.3
9.5
8.0
6.2
4.1
16
18
20
25
2.8
2.4
3.0
35
45
55
Age of Drivers
3.8
65
75
79
Solution
• f(x) = 0.013x2 – 1.19x + 28.24
• 3 = 0.013x2 – 1.19x + 28.24
• 0.013x2 – 1.19x + 25.24 = 0
a = 0.013; b = -1.19; c = 25.24
•
-(-1.19) ±√((-1.19)2 – 4(0.013)(25.24))
x = -------------------------------------------2(0.013)
• x ≈ ((1.19) ± √(0.104)) / (0.26)
x ≈ ((1.19) ± 0.322) / (0.26)
• x ≈ 33; x ≈ 58
Your Turn
•
Solve the following using the quadratic
formula
1. 4x2 – 3x = -6
2. 3x2 = 8x - 7
Discriminant
• Given: ax2 + bx + c = 0, where a > 0.
•
-b ± √(b2 – 4ac)
x = ----------------------2a
• If (b2 – 4ac) >= 0, x are real numbers
If (b2 – 4ac) < 0, x are imaginary numbers.
• (b2 – 4ac) is called a discriminant.
• Thus,
– If (b2 – 4ac) >= 0, solution set is real numbers.
– If (b2 – 4ac) , 0, solution set is complex numbers.
Your Turn
•
Compute the discriminant and determine
the number and type of solutions.
1. x2 + 6x = -9
2. x2 + 2x + 9 = 0
8.3 Quadratic Function &
Their Graphs
Quadratic function
f(x) = x2 - x - 2
f(x)
x
Characteristics of Quadratic
Function Graph
• f(x) = ax2 + bx + c
– Shape is a parabola
– If a> 0, parabola opens upward
– If a < 0, parabola opens downward
– Vertex is the lowest point (when a > 0), and
the highest point (when a < 0)
– Axis of symmetry is the line through the vertex
which divides the parabola into two mirror
images.
To Sketch a Graph of Quadratic
Function
• Given: f(x) = a(x – h)2 + k
• Characteristics
– If a > 0, parabola opens upward
– Vertex is at (h, k)
– If h > 0, graph is shifted to right by h; if h < 0,
to the left
– If k > 0, graph is shifted up by k; if k < 0,
downward by k
– Axis of symmetry: x = h
– For x-intercepts, solve f(x) = 0
Graph of f(x) = -2(x – 3)2 + 8
f(x)
-7
-192
-6
-154
-5
-120
-4
-90
-3
-64
-2
-42
-1
-24
-0
-10
1
0
2
6
3
8
4
6
5
0
6
-10
7
-24
f(x) = -2(x - 3)^2 + 8
50
0
-50
f(x)
x
-7 -6 -5 -4 -3 -2 -1 -0 1 2 3 4 5 6 7 8 9 10 11
-100
-150
-200
Excel
-250
x
Graph of f(x) = -2(x – 3)2 + 8
• Graph the function: f(x) = -2(x – 3)2 + 8
• Solution:
– Parabola opens downward (a = -2)
– Vertex: (3, 8)
– X-intercepts:
-2(x – 3)2 + 8 = 0
(x – 3)2 = -8/-2
x – 3 = ± √(4)
x = 5; x = 1
– y-intercept
f(0) = -2(0-3)2 + 8
= -2(9) + 8
= -10
Graph of f(x) = (x + 3)2 + 1
x
f(x)
-7
17.0
-6
10.0
-5
5.0
-4
2.0
-3
1.0
-2
2.0
-1
5.0
0
10.0
1
17.0
40.0
2
26.0
20.0
3
37.0
4
50.0
5
65.0
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
6
82.0
x
7
101.0
f(x) = (x + 3)^2 + 1
120.0
100.0
f(x)
80.0
60.0
0.0
Graph of f(x) = (x + 3)2 + 1
• Graph the function: f(x) = (x + 3)2 + 1
• Solution:
– Parabola opens upward (a = 1)
– Vertex: (-3, 1)
– X-intercepts:
(x + 3)2 + 1 = 0
(x + 3)2 = -1
x–3=±i
x = 3 + i; x = 3 - i
(This means no x-intercepts)
– Y-intercept
f(0) = (3)2 + 1 = 10
Graphing f(x) = ax2 + bx + c
• A
• Solution:
f(x) = a(x2 + (b/a)x) + c
= a(x2 + (b/a)x + (b2/4a2)) + c – a(b2/4a2)
= a(x + (b/2a))2 + c – b2/(4a)
• Comparing to f(x) = a(x – h)2 + k,
h = -b/(2a); k = c – b2/(4a)
• f(x) = a(x – (-b/(2a))2 + (c – b2/(4a))
Graphing f(x) = 2x2 + 4x - 3
• a = 2; b = 4; c = -3
• f(x) = a(x – (-b/(2a))2 + (c – b2/(4a))
= 2(x – (-4/(4))2 + (-3 – 16/8)
= 2(x + 1)2 – 5
• Opens upward
• Vertex: (-1, -5)
• X-intercept:
-1 - √(10)/2, -1 + √(10)/2
• Y-intercept:
f(0) = -3
Graphing f(x) = 2x2 + 4x - 3
f(x) = 2x^2 + 4x - 3
140.0
120.0
100.0
f(x)
80.0
60.0
40.0
20.0
0.0
-20.0 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
x