3.1
Quadratic Functions and Models
Quadratic Functions
A quadratic function is of the form
f(x) = ax2 + bx + c,
where a, b, and c are real numbers, a ≠ 0.




The graph of a quadratic function is a parabola.
The domain of a quadratic function is all real
numbers.
These functions have a linear rate of change.

Vertex

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Axis of symmetry


The maximum or minimum point of a parabola
The vertical line passing through the vertex
Leading coefficient



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In a quadratic function is this “a” (the coefficient of x2).
When positive the graph opens up.
When negative the graph opens down.
Larger values of |a| result in a narrower parabola, smaller
values of |a| result in a wider parabola.
Vertex Form of a Quadratic Function

Vertex Form

The parabolic graph of f(x) = a(x – h)2 + k has
vertex (h,k).

Graph opens up when a > 0, down when a < 0.
Examples

Page 184 Identify f as being linear, quadratic,
or neither. If it is quadratic, identify the
leading coefficient and evaluate f(-2).




#2 f(x) = 1 – 2x + 3x2
#4 f(x) = (x2 + 1)2
#6 f(x) = 1/5 x2
Page 185 Identify the vertex and the leading
coefficient. Then write the equation as f(x) =
ax2 + bx + c


#18 f(x) = 5(x + 2)2 – 5
#20 f(x) = ½(x + 3)2 – 5
Finding the vertex

Vertex Formula

The vertex for the graph of f(x) = ax2 + bx + c with
a ≠ 0 is the point
 b  b  
 2a , f  2a  
 

Examples

Page 185 Use the vertex formula to
determine the vertex of the graph of f.

#26 f(x) = 2x2 – 2x + 1

#30 f(x) = -3x2 + x – 2
Completing the Square

y = x2 + 6x – 8
Examples

Page 185 Write the given equation in the
form f(x) = (x – h)2 + k.

#40 f(x) = x2 + 10x + 7

#50 f(x) = 6 + 5x – 10x2
Quadratic Regression on the Calculator

Enter data into List 1 and List 2






Choose Stat
Calc
5: quadreg
enter
enter
To have the data go directly into y =





Before you press the second enter
Choose vars
y-vars
function
y1
Examples

Page 187 #98
a) Make a scatterplot of the data.
b) Find the values for a, h and k. Graph f(x)
together with the data in the same viewing
rectangle.
c) Approximate the undetermined value(s) in the
table.
U.S. population in millions
Year
1800
1820
1840
1860
1870
1880
Population
5
10
17
31
?
50
Year
1900
1920
1940
1960
1980
2000
Population
76
106
132
178
226
?
Problem Solving

Page 186 #82

Match the physical situation with the graph of the
quadratic function that models it best.
Example
120 ft
20 ft
300 ft

Page 187 #102



The cables that support a suspension bridge,
such as the Golden Gate Bridge, can be modeled
by parabolas.
Suppose that a 300-foot long suspension bridge
has towers at its ends that are 120 feet tall, as
illustrated in the accompanying figure.
If the cable comes within 20 feet of the road in the
center of the bridge, find the quadratic function
that models the height of the cable above the road
a distance of x feet from the center of the bridge.
3.2
Quadratic Equations and Problem
Solving
Examples

Page 201

#2
x  9 x  10  8
2

#10
8 x 2  63  46 x
Quadratic formula

The solutions to the quadratic equation
ax2 + bx + c = 0, where a ≠ 0, are given by
b  b  4ac
2a
2
x=

#16
3( x  5)  6  0
2

#18
3 2 1
1
x  x 0
4
2
2
The Discriminant

The discriminant is used to determine the
number of real solutions to ax2 + bx + c =0.



If b2 – 4ac > 0, there are two real solutions.
If b2 – 4ac = 0, there is one real solution.
If b2 – 4ac < 0, there are no real solutions.
Examples

Page 202
a. Write the equation in standard form
b. Calculate the discriminant and determine the number of real solutions
c. Solve the equation.

#46

#58
8 x  2  14
2
4x  6  x

#60
2
x(5 x  3)  1
Solve graphically

Page 202 #42
2 x  4 x  1.595
2
Problem Solving

Page 203 #100

From 1984 to 1994 the cumulative number of
AIDS cases can be modeled by the equation
C ( x)  3034 x 2  14, 018 x  6400,
Where x represents years after 1984. Estimate the
year when 200,000 AIDS cases had been
diagnosed.

Page 204 #108

A rectangular pen for a pet is under construction
using 100 feet of fence.
a. Determine the dimension that result in an area of 576
square feet.
b. Find the dimensions that give the maximum area.
3.3
Quadratic Inequalities
Solving Quadratic Inequalities

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Write in Standard Form
Solve
Use the boundary numbers to test points
Use the table or graph to write your solution
Examples

Page 213 Solve each equation and inequality. Write the
solution set for each inequality in interval notation.

#12
a. x 2  8 x  12  0
2
b. x  8 x  12  0
c. x 2  8 x  12  0

#14
a. n  17  0
2
b. n  17  0
2
c. n  17  0
2
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

#16
2
a. 7 x  4 x  0
2
b. 7 x  4 x  0
c. 7 x 2  4 x  0
#18
2
a. x  2 x  1  0
2
x
 2x 1  0
b.
2
c. x  2 x  1  0
#22
2
2
x
 4x  3  0
a.
2
b. 2 x  4 x  3  0
2
c. 2 x  4 x  3  0
3.4
Transformations of Graphs
Shifting and Stretching

Graph

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y1 = x2
y2 = x2 + 3
y3 = x2 – 3
What pattern do you see?
Vertical Shifts
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Vertical Shifts
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g(x) = f(x) + a, shift graph up a units
g(x) = f(x) – a, shift graph down a units
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Graph
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y1 = x2
y2 = (x + 3)2
y3 = (x – 3)2
What pattern do you see?
Horizontal Shifts
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Horizontal Shifts
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g(x) = f(x + a), shift graph left a units
g(x) = f(x – a), shift graph right a units
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Graph
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y1 = x2
y2 = 3x2
y3 = 6x2
Stretching
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Vertical and Horizontal stretches:
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For a >0, the graph g(x) = af(x) stretches the
graph vertically by a factor of a.
For a >1, the graph g(x) = f(ax) compresses the
graph horizontally by a factor of a.
h(x) = f(x/a) compresses the graph horizontally by
a factor of a.

Graph


y1 = x2
y2 = -x2
Negative Coefficients

When you multiply by a negative it reflects
(flips) the graph over the x-axis.

Predict what will happen

y = -x2 + 3

y = -2(x + 5)2 - 3

f(x) = x2, af(x + b) + c
Examples

Page 229 Use the accompanying graph of
y = f(x) to sketch a graph of each equation.

#12
a. y = f(x + 1)
b. y = -f(x)
c. Y = 2f(x)

#14
a. y = f(x – 1) - 2
b. y = -f(x) + 1
c. y = f(1/2x)
Other Parent Graphs

y=x

y = |x|

y = x3

y = √x
Examples

Page 230 Use transformations for graphs to
sketch a graph of f.

#50 f ( x)  x  1

#52 f ( x) | x  4 |

#68 f ( x)  ( x)3  1