1. Module 4 Quadratic Functions and Equations
Rev. F07
MAC 1140
Module 4
Quadratic Functions and Equations

2. Learning Objectives Upon completing this module, you should be able to
Rev. F07
Learning Objectives
Upon completing this module, you should be able to
understand basic concepts about quadratic functions and their graphs.
complete the square and apply the vertex formula.
graph a quadratic function by hand.
solve applications and model data.
understand basic concepts about quadratic equations.
use factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equations.
understand the discriminant.
solve problems involving quadratic equations.
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3. Learning Objectives (Cont.)
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Learning Objectives (Cont.)
9. solve quadratic inequalities graphically.
solve quadratic inequalities symbolically.
graph functions using vertical and horizontal translations.
graph functions using stretching and shrinking.
graph functions using reflections.
combine transformations.
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4. Quadratic Functions and Equations
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Quadratic Functions and Equations
There are four sections in this module:
3.1 Quadratic Functions and Models
3.2 Quadratic Equations and Problem- Solving
3.3 Quadratic Inequalities
3.4 Transformations of Graphs
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5. Rev. F07
Let’s get started by looking at some basic concepts about quadratic functions and graphs.
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6. Rev. F07
Quadratic Functions
Recall that a linear function can be written as f(x) = ax + b (or f(x) = mx + b). The formula for a quadratic function is different from that of a linear function because it contains an x2 term.
f(x) = 3x2 + 3x + 5 g(x) = 5  x2
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7. Quadratic Functions (Cont.)
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Quadratic Functions (Cont.)
The graph of a quadratic function is a parabola—a U shaped graph that opens either upward or downward.
A parabola opens upward if a is positive and opens downward if a is negative.
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8. Quadratic Functions (Cont.)
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Quadratic Functions (Cont.)
The highest point on a parabola that opens downward and the lowest point on a parabola that opens upward is called the vertex.
The vertical line passing through the vertex is called the axis of symmetry.
The leading coefficient a controls the width of the parabola. Larger values of |a| result in a narrower parabola, and smaller values of |a| result in a wider parabola.
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9. Example of Different Parabolas
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Example of Different Parabolas
Note: A parabola opens upward if a is positive and opens downward if a is negative.
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10. Example of Different Parabolas (Cont.)
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Example of Different Parabolas (Cont.)
Note: The leading coefficient a controls the width of the parabola. Larger values of |a| result in a narrower parabola, and smaller values of |a| result in a wider parabola.
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11. Graph of the Quadratic Function
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Graph of the Quadratic Function
Now, let’s use the graph of the quadratic function shown to determine the sign of the leading coefficient, its vertex, and the equation of the axis of symmetry.
Leading coefficient: The graph opens downward, so the leading coefficient a is negative.
Vertex: The vertex is the highest point on the graph and is located at (1, 3).
Axis of symmetry: Vertical line through the vertex with equation
x = 1.
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12. Let’s Look at the Vertex Form of a Quadratic Function
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Let’s Look at the Vertex Form of a Quadratic Function
We can write the formula f(x) = x2 + 10x + 23 in vertex form by completing the square.
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13. Rev. F07
Let’s Write the Vertex Form of a Quadratic Function by Completing the Square
Subtract 23 from each side.
Let k = 10; add (10/2)2 = 25.
Factor perfect square trinomial.
Vertex Form
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14. Rev. F07
How to Use Vertex Formula to Write a Quadratic Function in Vertex Form?
Use the vertex formula to write f(x) = 3x2  3x + 1 in vertex form.
Solution: Since a = -3, b = -3 and c = 1, we just need to substitute them into the vertex formula. Mainly, you need to know the vertex formula for x; once you have solved for x, you can solve for y.
1. Begin by finding Find y.
the vertex.
The vertex is:
Vertex form:
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15. Example Graph the quadratic equation g(x) = 3x2 + 24x  49. Solution
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Example
Graph the quadratic equation g(x) = 3x2 + 24x  49.
Solution
The formula is not in vertex form, but we can find
the vertex.
The y-coordinate of the vertex is:
The vertex is at (4, 1). The axis of symmetry is x = 4, and the parabola opens downward because the leading coefficient is negative.
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16. Example (Cont.) x y 2 13 3 4 4 1 5 6 vertex
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Example (Cont.)
Graph: g(x) = 3x2 + 24x  49
Table of Values x y 2
13 3 4 4 1 5 6
vertex
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17. Example of Application
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Example of Application
A junior horticulture class decides to enclose a rectangular garden, using a side of the greenhouse as one side of the rectangle. If the class has 32 feet of fence, find the dimensions of the rectangle that give the maximum area for the garden. (Think about using the vertex formula.)
Solution
Let w be the width and L be the
length of the rectangle.
Because the 32-foot fence does not go along the greenhouse, if follows that W + L + W = or L = 32 – 2W L W
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18. Example of Application (Cont.)
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Example of Application (Cont.)
The area of the garden is the length times the width.
This is a parabola that opens downward, and by the vertex formula, the maximum area occurs when
The corresponding length is L = 32 – 2W = 32 – 2(8) = 16 feet.
The dimensions are 8 feet by 16 feet. L W
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19. Rev. F07
Another Example
A model rocket is launched with an initial velocity of vo = 150 feet per second and leaves the platform with an initial height of ho = 10 feet.
a) Write a formula s(t) that models the height of the rocket after t seconds.
b) How high is the rocket after 3 seconds?
c) Find the maximum height of the rocket Support your answer graphically.
Solution a)
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20. Another Example (Cont.)
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Another Example (Cont.) b)
The rocket is 316 feet high after 3 seconds.
c) Because a is negative, the vertex is the highest point on the graph, with an t-coordinate of
The y-coordinate is:
The vertex is at (4.7, 361.6).
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21. How to Solve Quadratic Equations?
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How to Solve Quadratic Equations?
The are four basic symbolic strategies in which quadratic equations can be solved.
Factoring
Square root property
Completing the square
Quadratic formula
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22. Rev. F07
Factoring
A common technique used to solve equations that is based on the zero-product property.
Example
Solution
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23. Square Root Property Example: Rev.S08 Rev. F07
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24. Rev. F07
Completing the Square
Completing the square is useful when solving quadratic equations that do not factor easily.
If a quadratic equation can be written in the form where k and d are constants, then the equation can be solved using
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25. Completing the Square (Cont.)
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Completing the Square (Cont.)
Solve 2x2 + 6x = 7.
Solution
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26. Quadratic Formula Solve the equation Solution
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Quadratic Formula
Solve the equation
Solution
Let a = 2, b = 5, and c = 9.
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27. Quadratic Equations and the Discriminant
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Quadratic Equations and the Discriminant
Use the discriminant to determine the number of solutions to the quadratic equation
Solution
Since , the equation has two real solutions.
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28. Example of a Modeling a Projectile Motion
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Example of a Modeling a Projectile Motion
The following table shows the height of a toy rocket launched in the air.
Use to model the data.
After how many seconds did the toy rocket strike the ground?
Height of a toy rocket
t (sec) 1 2
s(t) feet 12 36 28
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29. Example (Cont.) If t = 0, then s(0) = 12, so
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Example (Cont.)
If t = 0, then s(0) = 12, so
The value of vo can be found by noting that when t = 2, s(2) = 28. Substituting gives the following result.
Thus s(t) = 16t2 + 40t + 12 models the height of the toy rocket.
The rocket strikes the ground when
s(t) = 0, or when –16t2 + 40t + 12 = 0.
Using the quadratic formula, where a = 4,
b = – 10 and c = – 3 we find that
Only the positive solution is possible, so the toy rocket reaches the ground after approximately 2.8 seconds.
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30. Rev. F07
One More Example
A box is is being constructed by cutting 2 inch squares from the corners of a rectangular sheet of metal that is 10 inches longer than it is wide. If the box has a volume of 238 cubic inches, find the dimensions of the metal sheet.
Solution
Step 1: Let x be the width and x + 10 be the length.
Step 2: Draw a picture.
Since the height times the width times the length must equal the volume, or 238 cubic inches, the following can be written x
x - 4
x + 6
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31. One More Example (Cont.)
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One More Example (Cont.)
Step 3: Write the quadratic equation in the form ax2 + bx + c = 0 and factor.
The dimensions can not be negative, so the width is 11 inches and the length is 10 inches more, or 21 inches.
Step 4: After the 2 square inch pieces are cut out, the dimensions of the bottom of the box are 11 – 4 = 7 inches by 21 – 4 = 17 inches. The volume of the box is then 2 x 7 x 17 = 238, which checks.
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32. Rev. F07
Quadratic Inequality
If an equals sign is replaced by >, , <, or ≤, a quadratic inequality results.
A first step in solving a quadratic inequality is to determine the x-values where equality occurs.
These x-values are the boundary numbers.
Let’s look at a quadratic inequality graphically.
The graph of a quadratic function opens
either upward or downward. In this case,
a = 1, parabola opens up, and we have
x-intercepts: 1 and 2
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33. Quadratic Inequality (cont.)
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Quadratic Inequality (cont.)
Note the parabola lies below the x-axis
between the intercepts for the
equation x2  x  2 = 0
Solutions to x2  x  2 < 0, is the solution set {x|1 < x < 2} or (1, 2) in interval notation.
Solutions to x2  x  2 > 0 include
x-values either left of x = 1 or right of x = 2, where the parabola is above the x-axis, and thus
{x|x < 1 or x > 2}.
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34. Rev. F07
Another Example
Solve the inequality. Write the solution set for each in interval notation.
a) 3x2 + x  4 = 0 b) 3x2 + x  4 < 0
c) 3x2 + x  4 > 0
Solution
a) Factoring (3x + 4)(x – 1) = 0
x = – 4/3 x = 1
The solutions are – 4/3 and 1.
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35. Another Example (Cont.)
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Another Example (Cont.)
y < 0
y > 0
b) 3x2 + x  4 < 0
Parabola opening upward.
x-intercepts are 4/3 and 1
Below the x-axis (y < 0)
Solution set: (– 4/3, 1)
c) 3x2 + x  4 > 0
Above the x axis (y > 0)
Solution set:
(, 4/3)  (1, )
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36. Solving Quadratic Inequalities in 4 Steps
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Solving Quadratic Inequalities in 4 Steps
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37. Transformation of Graphs: Vertical Shifts
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Transformation of Graphs: Vertical Shifts
A graph is shifted up or down. The shape of the graph is not changed—only its position.
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38. Transformation of Graphs: Horizontal Shifts
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Transformation of Graphs: Horizontal Shifts
A graph is shifted left or right. The shape of the graph is not changed—only its position.
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39. Transformation of Graphs
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Transformation of Graphs
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40. Example of Transformation of Graphs
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Example of Transformation of Graphs
Shifts can be combined to translate a graph of y = f(x) both vertically and horizontally.
Shift the graph of y = x2 to the left 3 units and downward 2 units.
y = x y = (x + 3) y = (x + 3)2  2
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41. Rev. F07
Another Example
Find an equation that shifts the graph of f(x) = x2  2x + 3 left 4 units and down 3 units.
Solution
To shift the graph left 4 units, replace x with (x + 4) in the formula for f(x).
y = f(x + 4) = (x + 4)2 – 2(x + 4) + 3
To shift the graph down 3 units,
subtract 3 to the formula.
y = f(x + 4)  3 = (x + 4)2 – 2(x + 4) + 3  3
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42. Vertical Stretching and Shrinking
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Vertical Stretching and Shrinking
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43. Horizontal Stretching and Shrinking
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Horizontal Stretching and Shrinking
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44. Reflection of Graphs Rev.S08 Rev. F07
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45. Rev. F07
Reflection of Graphs
For the function f(x) = x2 + x  2 graph its reflection across the x-axis and across the y-axis.
Solution The graph is a parabola with x-intercepts 2 and 1.
To obtain its reflection across the x-axis, graph y = f(x), or y = (x2 + x  2). The x-intercepts have not changed.
To obtain the reflection across the y-axis let y = f(x), or
y = (x)2  x  2. The x-intercepts have changed to 1 and 2 .
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46. Combining Transformations
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Combining Transformations
Transformations of graphs can be combined to create new graphs.
For example the graph of y = 3(x + 3)2 + 1 can be obtained by performing four transformations on the graph of y = x2.
1. Shift of the graph 3 units left: y = (x + 3)2
2. Vertically stretch the graph by a factor of 3: y = 3(x + 3)2
3. Reflect the graph across the x-axis:
y = 3(x + 3)2
4. Shift the graph upward 1 unit: y = 3(x + 3)2 + 1
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47. Combining Transformations (Cont.)
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Combining Transformations (Cont.)
y = 3(x + 3)2 + 1
Shift to the left 3 units.
Shift upward 1 unit.
Reflect across the x-axis.
Stretch vertically by a factor of 3
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48. What have we learned? We have learned to
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What have we learned?
We have learned to
understand basic concepts about quadratic functions and their graphs.
complete the square and apply the vertex formula.
graph a quadratic function by hand.
solve applications and model data.
understand basic concepts about quadratic equations.
use factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equations.
understand the discriminant.
solve problems involving quadratic equations.
Click link to download other modules.
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49. What have we learned? (Cont.)
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What have we learned? (Cont.)
9. solve quadratic inequalities graphically.
solve quadratic inequalities symbolically.
graph functions using vertical and horizontal translations.
graph functions using stretching and shrinking.
graph functions using reflections.
combine transformations.
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50. Rev. F07
Credit
Some of these slides have been adapted/modified in part/whole from the slides of the following textbook:
Rockswold, Gary, Precalculus with Modeling and Visualization, 3th Edition
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