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QUADRATIC FUNCTIONS
Monika V Sikand
Light and Life Laboratory
Department of Physics and Engineering physics
Stevens Institute of TechnologyHoboken, New Jersey, 07030.
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OUTLINE
Graphing Quadratic Functions
Solving Quadratic Equations by Factoring
Solving Quadratic Equations by FindingSquare Roots
Complex Numbers
The Quadratic Formula and the DiscriminantGraphing and Solving Quadratic Inequalities
Modeling with Quadratic Functions
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Graphing Quadratic Functions
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THE GRAPH OF A QUADRATIC
FUNCTION
vertex
Axis of
symmetry
y = x2
y = -x2
The parabola opens upif a>0 andopens downif a 1.
The x-coordinate of the vertexis-b/2a.
The axis of symmetryis the vertical
line x = -b/2a.
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EXAMPLE
Graph y = 2x2-8x +6
Solution: The coefficients for this function are
a = 2, b = -8, c = 6.
Since a>0, the parabola opens up.
The x-coordinate is: x = -b/2a, x = -(-8)/2(2)
x = 2
The y-coordinate is: y = 2(2)2-8(2)+6
y = -2
Hence, the vertex is (2,-2).
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EXAMPLE(contd.)
Draw the vertex (2,-2)on graph.
Draw the axis of symmetry x=-b/2a.
Draw two points on one side of the
axis of symmetry such as (1,0) and
(0,6).
Use symmetry to plot two more points
such as (3,0), (4,6).
Draw parabola through the plotted
points.(2,-2)
(1,0)
(0,6)
(3,0)
(4,6)
Axis of symmetry
x
y
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VERTEX FORM OF QUADRATIC
EQUATION
y = a(x - h)2+ k
The vertex is (h,k).
The axis of symmetry is x = h.
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GRAPHING A QUADRATIC
FUNCTION IN VERTEX FORM
(-3,4)
(-7,-4)
(-1,2)(-5,2)
(1,-4)
Axis of symmetry
x
y
Exampley = -1/2(x + 3)2+ 4
where a = -1/2, h = -3, k = 4. Since a
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INTERCEPT FORM OF
QUADRATIC EQUATION
y = a(x - p)(x - q)
The x intercepts are p and q.
The axis of symmetry is halfway
between (p,0) and (q,0).
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GRAPHING A QUADRATIC
FUNCTION IN INTERCEPT FORM
(-2,0)
(1,9)
(4,0)
Axis of symmetry
x
y
Exampley = -(x + 2)(x - 4).
where a = -1, p = -2, q = 4. Since
a
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WRITING THE QUADRATIC
EQUATION IN STANDARD FORM(1). y = -(x + 4)(x - 9)
= -(x2 - 9x + 4x - 36)
= -(x2 - 5x -36)
= -x2 + 5x + 36
(2). y = 3(x -1)2+ 8
= 3(x -1)(x - 1) + 8
= 3(x2- x - x + 1) + 8= 3(x2- 2x + 1) + 8
= 3x2- 6x + 3 + 8
= 3x2- 6x + 11
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SOLUTION
Since a = -3.678 is negative, the graph of the quadratic function open
down and the function has a maximum value. The maximum value
occurs at:
The corresponding value of y is:
a. Hence, The temperature that made the greatest percent of test
subjects comfortable was about 72.
b. At that temperature about 92% of the subjects felt comfortable.
x b2a
527.32(3.678)
72
y 3.678(72)2
527.3(72)18,807 92
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REAL LIFE EXAMPLE 6
(page 252)The Golden Gate Bridge in San Franciscohas two towers that
rise 500 feet above the road and are connected by suspension
cables as shown. Each cable forms a parabola with equation
where x and y are measured in feet.
a. What is the distance d between the towers?
b. What is the height l above the road of a cable at its lowest
point?
y 1
8940(x 2100)2 8
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GOLDEN GATE BRIDGE
l
d
y
200ft
500ft
x
Not down to scale
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SOLUTION
y 1
8960(x 2100)
2 8
y 1
8960 (x2
4200x 4410000) 8
y 1
8960x2
4200
8960x (
4410000
8960 8)
x b
2a
(4200
8960
)( 1
2 18960
)
x 2100
y 1
8960(x 2100)2 8
y 1
8960 (2100 2100)2
8
y 0
8960 8
y 8
Hence the vertex of the parabola is (2100,8)
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SOLUTION(contd.)
a. The vertex of the parabola is (2100,8), so the
cables lowest point is 2100 feet from the left tower
shown above. Since the heights of the two towersare the same, the symmetry of the parabola implies
that the vertex is also 2100 feet from the right tower.
Therefore the towers are d = 2(2100) = 4200 feet
apart.
b. The height l above the road of a cable at its lowest
point is the y-coordinate of the vertex. Since the
vertex is (2100,8), this height is l= 8 feet.
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SOLVING QUADRATICEQUATION BY FACTORING
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FACTORING QUADRATIC
EXPRESSIONThe expression x2+bx+ cis a trinomial because it
has three terms. We can use factoring to write it as
product of two terms or binomials such as
x2+bx+ c = (x+m)(x+n) = x2 + (m+n)x + mn
Example:
x2+ 8x + 15 = (x + 3)(x + 5)
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EXAMPLE
Problem: Factor x2- 12x - 28
Solution:
x2- 12x - 28 = (x+m)(x+n) where mn = -28 and m+n = -12
Factors
of -28
-1,28 1,-28 -2,14 2,-14 -4,7 4,-7
Sum of
factors
27 -27 12 -12 3 -3
The table shows that m = 2 and n = -14.
So, x2- 12x - 28 = (x + 2)(x - 14)
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SPECIAL FACTORING
PATTERNS1. Difference of two squares: a2- b2= (a+b)(a-b)
Example: x2
- 9 = (x+3)(x-3)
2. Perfect square Trinomial: a2+ 2ab + b2= (a+b)2
Example: x2+ 12x +36 = (x+6)2
3. Perfect square Trinomial: a2- 2ab + b2= (a-b)2
Example: x2-8x +16 = (x-4)2
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FACTORING MONOMIALS FIRST
Monomialis an expression that has only one term.
Factor the quadratic expression:
a. 5x2- 20 = 5(x2-4)
= 5(x+2)(x-2)
b. 6p2 + 15p + 9 = 3(2p2+ 5p + 3)
= 3(2p + 3)(p + 1)
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SOLVING QUADRATIC
EQUATIONS
x2 3x 18 0
(x 6)(x 3) 0
x 6 0
x 6
or
x 3 0,
x 3
Solve: Solve:2t2 17t 45 3t 5
2t2 20t 50 0
t2 10t 25 0
(t 5)2 0
t 5 0
t 5
The solution is 5.
The solutions are -6 and 3.
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REAL LIFE EXAMPLE
You have made a rectangular stained glass window
that is 2 feet by 4 feet. You have 7 square feet of clear
glass to create a border of uniform width around the
window. What should the width of the border be?
x xx x
x
x xx
2
44+2x
2+2x
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ZERO PRODUCT PROPERTY
Let A and B be real numbers or algebraic
expressions. If AB=0, then either A = 0 or B = 0
Solve(a) x2+ 3x -18 = 0
(x + 6)(x - 3) = 0.Hence either x + 6 = 0 or x - 3 = 0.
The solutions are x = -6 or x = 3
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ZERO PRODUCT PROPERTY
Let A and B be real numbers or algebraic
expressions. If AB=0, then either A = 0 or B = 0
Solve(a) 2t2- 17t + 45 = 3t - 5
2t2- 20t + 50 = 0
t2- 10t + 25 = 0
(t - 5)2
= 0t - 5 = 0
t = 5
Hence the solutions is 5.
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FINDING ZEROES OF QUADRATIC
FUNCTIONS
Find zeros of y = x2-x -6
Solution:
y = x2 - 3x + 2x - (32)
y = x(x-3) + 2(x-3)
y = (x + 2)(x - 3)
Hence the zeros of the function are -2 and 3.
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SOLVING QUADRATICEQUATIONS BY FINDING
SQUARE ROOTS
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SQUARE ROOT
A number r is a square rootof a number s if r2= s.
A positive number s has square roots denoted by
Example:
Since 32= 9 and (-3)2= 9. The square roots of 9 are:
sor s
9 3
9 3.
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PROPERTIES OF SQUARE ROOTS
(a>0, b>0)
PRODUCT PROPERTY:
QUOTIENT PROPERTY:
ab a. b
a
b
a
b
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SIMPLIFY
(a). 24
4 . 6
2 6
(b). 7
2
7
2
2
2
14
2
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Solving a quadratic equation
2x2 117
2x2
16x2 8
x 8
x2 2
Hence the solutions are:
2 2 and 2 2
1
3
(x 5)2 7
(x 5)2 21
(x 5) 21
x 5 21
5 21 5 21
Hence the solutions are:
and
Solve: Solve:
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REAL LIFE EXAMPLE
A stunt man working on the set of a movie is to fall
out of a window 100 feet above the ground. For the
stunt mans safety, an air cushion 26 feet wide by
30 feet long by 9 feet high is positioned on theground below the window.
a. For how many seconds will the stunt man fall before
he reaches the cushion?b. A movie camera operating at a speed of 24 frames
per second records the stunt mans fall. How many
frames of film show the stunt man falling?
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SOLUTION
a. The stunt mansinitial height is ho= 100 feet, so his height as a
function of time can be modeled by function h = -16t2 + 100.
Since the height of the cushion is 9 feet above the ground, the
the time taken by the stunt man to reach the cushion is:h = -16t2 + 100
9 = -16t2 + 100
-91 = -16t2
91/16 = t2
or t 2.4. Thus, it takes about 2.4 seconds for the stuntman to reach the cushion.
b. The number of frames of filmthat show the stunt man falling is
given by the product (2.4sec)(24 frames/sec), or about 57
frames.
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COMPLEX NUMBERS
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COMPLEX NUMBER
A complex number written in standard form is a
number a+bi where a and b are real numbers.
The number a is the real part of the complex
number and number bi is the imaginary part.
If b0, then a+bi is an imaginary number. If a=0 and
b0, then a+bi is a pure imaginary number.
A complex plane has a horizontal axis called the
real axisand a vertical axis called the imaginary axis.
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THE SQUARE ROOT OF A
NEGATIVE NUMBER
1. If r is a positive real number, then
whereiis the imaginary unitdefined as
2. By property (1), it follows that
r i r
(i r)2 r
i 1
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SOLVING A QUADRATIC
EQUATION
3x2 10 26
Solution
3x2 10 26
3x2 36
x2
12
x 12
x i 12
x 2i 3
Hence, the solutions are
and 2i 3
2i 3
Solve:
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PLOTTING COMPLEX NUMBERS
Plot 2-3i in the complex
plane.
To plot 2-3i , start at theorigin, move 2 units to the
right and then move 3 units
down.
0
2-3i
x
y
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ADDING AND SUBTRACTING
COMPLEX NUMBERS
Sum of complex numbers:
(a+bi) + (c+di) = (a+c) + i(b+d)
Example:
(4-i) + (3+2i) = (4+3) + i(-1+2)
= 7 + i
Difference of complex numbers:
(a+bi) - (c+di) = (a-c) + i(b-d)
Example:
(7-5i) - (1-5i) = (7-1) + i(-5+5)
= 6 + 0i
= 6
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MULTIPLYING THE COMPLEX
NUMBERS
Write the expression as a complex number in standard form.
a. 5i(-2+i)= -10i + 5i2
= -10i + 5(-1)
= -5-10i
b. (7-4i)(-1+2i)= -7 + 14i + 4i - 8i2
= -7 + 18i - 8(-1)= 1 + 18i
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DIVIDING COMPLEX NUMBERS
Write the quotient in standard form.
Solution:Multiply the numerator and denominator by the complex
conjugate of the denominator.
5 3i
1 2i
5 3i
1 2i 1 2i
1 2i 5 10i 3i 6i2
1 2i 2i 4i2
113i
5
1
5
13
5i
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ABSOLUTE VALUES OF COMPLEX
NUMBER
Find the absolute value of each complex number.
(a)3 4i 32 42 25 5
(b)2i 0 (2i) 02
(2)2
2
(c)1 5i (1)2 52 265.10
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THE QUADRATIC FORMULA
Let a, b, and c be real numbers such that a0. The
solutions of the quadratic equation ax2+ bx +c = 0
are:
x b b
2 4ac
2a
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SOLVING QUADRATIC EQUATION
WITH TWO REAL SOLUTIONS
Solve
2x2 x 5
Solution:
2x2x 5 0
xb b
2 4ac
2a
x 1 12
4(2)(5)2(2)
x1 41
4
x 1 41
4 1.35
and
x 1 41
4 1.85
The solutions are:
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SOLVING QUADRATIC EQUATION
WITH ONE REAL SOLUTIONS
Solve
x2 x 5x 9
Solution:x2 x 5x 9
x2 6x 9 0
x6 (6)2 4(1)(9)
2(1)
x6 0
2
x3
Hence, the solution is 3.
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SOLVING QUADRATIC EQUATION
WITH TWO IMAGINARY SOLUTIONS
Solve
x2
2x 2
Solution
x2 2x 2
x2
2x 2 0
x2 22 4(1)(2)
2(1)
x
2 4
2
x2 2i
2
x1 i
The solutions are:
1+i and 1-i
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DISCRIMINANT
In the quadratic formula, the expression b2-4ac
under the radical sign is called the discriminant
of the associated equation ax2
+ bx + c = 0.
x
b b2 4ac
2a
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EXAMPLE: TWO REAL SOLUTIONS
Solve
x2 6x 8 0
Discriminant:
b2 4ac (6)2 4(1)(8) 4
Solutions:
xb b2 4ac
2a
x 6 (6)2 4(1)(8)2
x6 4
2
6 2
2314, 2
Hence there are two realsolutions: 4,2
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EXAMPLE: ONE REAL SOLUTION
Solve
x2 6x 9 0
Discriminant:
b2 4ac (6)2 4(1)(9) 0
Solutions:
xb b2 4ac
2a
x 6 (6)2
4(1)(9)2
x6 0
2
6
23
Hence, there is one real solution: 3
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EXAMPLE: TWO IMAGINARY
SOLUTIONS
Find the discriminant of the quadratic equation and give the number
and type of solutions of the equation.
Solve
x2
6x10 0
Discriminant:
b2
4ac (6)2
4(1)(10) 4
Solutions:
x b b
2 4ac
2a
x 6 (6)
2 4(1)(10)
2
x 6 4
262 1
2 3 i
Hence there are twoimaginary solutions:
3+i and 3-i
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SOLUTION(contd.)
h 16t2 vot ho
5 16t2 45t 6
0 16t2
45t1
t45 2089
32
t 0.022
ort 2.8
Rejecting the negative solution, the
baton is in the air for about 2.8seconds.
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GRAPHING AND SOLVINGQUADRATIC INEQUALITIES
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GRAPHING A QUADRATIC
INEQUALITY IN TWO VARIABLES
y ax2+bx +c y ax2+bx +c
Draw parabola with equation y = ax2+bx +c . Makethe parabola dashed for inequalities with < or >and
solid for inequalities with or .
Choose a point (x,y) inside the parabola and check
whether the point is a solution of the inequality.
If a point (x,y) is a solution, shade the region inside
the parabola. If it is not the solution, shade the region
outside the parabola.
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EXAMPLE
0(1,0) x
y
4
1
Graph y > x2-2x -3
Solution:
Graph y = x2-2x -3. Since theinequality is >, make parabola dashed.
Test the point (1,0) inside the
parabola. y > x2
-2x -3
0 > 12-2(1)-3
0 > -4So, (1,0) is a solution of the inequality.
Shade the region inside the parabola.
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REAL LIFE EXAMPLE
You are building a wooden bookcase. You want
to choose a thickness d(in inches) for the
shelves so that each is strong enough to support
60 pounds of books without breaking. A shelf cansafely support a weight of W ( in pounds)
provided that
W 300d2
a. Graph the given inequality
b. If you make each shelf 0.75 inch thick, can it
support a weight of 60 pounds?
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SOLUTION
y
0
50
100
150
200
250300
W
d0.5 1.0 1.5
a. Graph W = 300d2 for non negative values of
d. Since the inequality symbol is make the
parabola solid. Test a point inside the
parabola, such as (0.5, 240).
W 300d2
240 300(0.5)2240 75
Since the chosen point is not a solution,
shade the region outside (below) the
parabola.
b. The point (0.75,60) lies in the shaded regionof the graph from part (a), so (0.75,60) is a
solution of the given inequality.
c. Therefore, a shelf that is 0.75 inch thick can
support a weight of 60 pounds.
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GRAPHING A SYSTEM OF
QUADRATIC INEQUALITY
x
yGraph the system of quadratic inequalities.
yx2-4 Inequality 1
y
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QUADRATIC INEQUALITY IN
ONE VARIABLE
1. To solve ax2+ bx + c < 0 (or ax2 + bx + c 0),
graph y = ax2+ bx + c and identify the x values
for which the graph lies below (or on and
below) the x-axis.
2. To solve ax2+ bx + c > 0 (or ax2 + bx + c 0),
graph y = ax2+ bx + c and identify the x values
for which the graph lies above(or on andbelow)the x-axis.
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EXAMPLE
x
y
3
1
1 5
Solve x2- 6x + 5 < 0
Solution:
1. The solution consist of the x-values for
which the graph of y = x2 - 6x + 5 lies
below the x-axis.2. Find the graphs x-intercepts by letting y
= 0
0 = x2 - 6x + 5
0 = (x-1)(x-5)
x = 1 or x = 5
Sketch a parabola that opens up and has1 and 5 as x-intercepts.
3. The graph lies below the x-axis between
x = 1 and x = 5.
4. The solution of the given inequality is
1
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EXAMPLE
Solve 2x2 + 3x -3 0
Solution:
1. The solution consist of the x-values for which the graph ofy = 2x2 + 3x -3 lies on and above the x-axis.
2. Find the graphs x-intercepts by letting y = 0
0 = 2x2 + 3x -3
x3 32 4(2)(3)
2(2)
x3 33
4
x 0.69or
x 2.19
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EXAMPLE(contd.)
-2.19 4 x
y
0.69
1
1. Sketch a parabola that opens up
and has 0.69 and -2.19 as x-
intercepts.
2. The graph lies on and above the x-
axisto the left of ( and including) x
= -2.19 and to the right of ( and
including) x = 0.69.
3. The solution of the given inequality
is approximately x -2.19 or x
0.69.
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SOLVING A QUADRATIC
INEQUALITY ALGEBRAICALLY
Solve:
x2+ 2x 8
Solution:
First replace the inequality symbol with equal sign.x2+ 2x = 8
x2+ 2x - 8 = 0
(x+4)(x-2) = 0
x = - 4 or x = 2
The numbers -4 and 2 are the critical x-valuesof the inequality x2+ 2x
8. Plot -4 and 2 on a number line.
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SOLUTION (contd.)
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
These critical x-values partition the number line into
three intervals.
Test x = -5, (-5)2+ 2(-5) = 15 8
Test x = 0, (0)2 +2(0) = 0 8Test x = -3, (3)2+ 2(3) = 15 8
Hence the solution is -4 x 2.
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REAL LIFE EXAMPLE
For a driver aged x years, a study found that the drivers reaction time V(x)
(in milliseconds) to a visual stimulus such as traffic can be modeled by:
V(x) = 0.005x2 - 0.23x + 22, 16 x 70
At what ages does a drivers reaction time tend to be greater than 25
milliseconds?
Solution:
The values of x for which
V(x) > 25
0.005x2- 0.23x +22 > 25
0.005x2- 0.23x - 3 > 0The solution consists of the x- values for which the graph lies above the x-
axis.
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SOLUTION(contd.)
The graphs x intercept is found by letting y =0 and using the quadratic
formula to solve for x.
0.005x2- 0.23x - 3 = 0
x 0.23 (0.23)2 4 0.005 (3)
2 0.005
x0.23 .0529 .06
.01
x0.23 .1129
.01
x
0.23 0.34
.01
x 57approx.
or
x 11approx
Rejecting the negative value, thegraphs x-intercept is about 57. The
graph of 0.005x2 - 0.23x - 3 = 0 lies in
the domain 16 x 70. The graph lies
above the x-axis when 57 < x 70.
Hence the drivers over 57 years old
tend to have reaction times greater than
25 milliseconds.
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MODELING WITHQUADRATIC FUNCTIONS
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QUADRATIC FUNCTION IN
VERTEX FORM
(2,-3)
(4,1)
x
y
1
1
Write the quadratic function for the parabola
shown.
Solution:
The vertex shown is (h,k) = (2,-3)Using the vertex form of the quadratic function.
y = a(x-h)2+ k
y = a(x-2)2- 3
Use the other given point (4,1) to find a.
1 = a(4-2)2- 3
1 = 4a - 3
4 = 4a1 = a
Hence the quadratic function for the parabola is
y = (x-2)2-3
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QUADRATIC FUNCTION IN
INTERCEPT FORM
(-1,2)
(-2)
(3)
x
y
1
1
Write the quadratic function for the parabola
shown.
Solution:
The x intercepts shown are p = -2, q = 3Using the intercept form of the quadratic
function.
y = a(x-p)(x-q)
y = a(x+2)(x-3)
Use the other given point (-1,2) to find a.
2 = a(-1+2)(-1-3)
2 = -4a-1/2 = a
Hence the quadratic function for the parabola
is y = -1/2(x+2)(x-3)
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END