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Types of functions

Polynomial

Linear (chapter 2)
Non-linear

Degree 2 (chapter 3)
Degree >2 (chapter 4 part 1)

Rational (chapter 4 part 2)
Exponential (chapter 5)
Logarithmic (chapter 5)
Trigonometric (chapters 6-8)

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Major parts of quadratic

Vertex (maximum and minimum)
x-intercepts (zeros)

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Quadratic Functions �and Equations

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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Quadratic Functions and Models

Learn 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.

3.1

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Basic Concepts�

Linear function

f(x) = ax + b (or f(x) = mx + b).

Quadratic function contains an x² term in the numerator.

f(x) = 3x² + 3x + 5 g(x) = 5 − x²

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Identify Quadratic Function

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Quadratic Function�

f(x)=ax²+bx+c
parabola

opens up +a
opens down -a

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Quadratic Function�

f(x)=ax²+bx+c
parabola

a controls the width parabola

Larger |a| result in a narrower parabola
Smaller |a| result in a wider parabola

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Quadratic Function�

f(x)=ax²+bx+c
parabola

Vertex highest point or lowest point

Maximum or Minimum

Axis of symmetry vertical line passing through the vertex.

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Example�

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.

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Solution

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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Example Find the vertex of the graph of

symbolically. Support your

answer graphically.

a = 1/2 , b = −4, and c = 8.

x-coordinate of vertex:

y-coordinate evaluate f(4):

The vertex is (4, 0).

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Example�

Graph the quadratic equation g(x) = −3x² + 24x − 49.

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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Example�

Use the vertex formula to write f(x) = −3x² − 3x + 1 in vertex form.

1. Begin by finding 2. Find y.

the vertex.

The vertex is:

Vertex form:

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Example�

Write the formulas in vertex form by completing the square.

f(x) = x² + 10x + 23
g(x) = x² + 6x – 3

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Example�

Write the formulas in vertex form by completing the square.

f(x) = x² + 10x + 23
g(x) = x² + 6x – 3
j(x) = 2x² - 5x + 15
h(x) = (1/3)x² - x + 2
k(x) = ax² + bx + c

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Applications and Models Example�

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.

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 = 32 or L = 32 – 2W

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L

W

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Example�

A model rocket is launched with an initial velocity of v_o = 150 feet per second and leaves the platform with an initial height of h_o = 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.

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Example�

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. If the cable comes within 20 feet of the road in the center of the bridge, find a quadratic function that models the height of the cable above the road a distance of x feet from the center of the bridge.

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120 ft

20 ft

300 ft

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Quadratic Equations and Problem Solving

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

3.2

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Major parts of quadratic

Vertex (maximum and minimum)
x-intercepts (zeros)

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Solving Quadratic Equations�

Finding the x-intercepts
Finding the zeros

Four symbolic strategies:

Factoring (special cases)
Square root property (special cases)
Completing the square (always works)
Quadratic formula (always works)

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Factoring�

Based on the zero-product property.

Examples

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Square Root Property

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Example�

A rescue helicopter hovers 68 feet above a jet ski in distress and drops a life raft. The height in feet of the raft above the water is given by

Determine how long it will take for the raft to hit the water after being dropped from the helicopter.

Solution (continued on next slide)

The raft will hit the water when its height is 0 feet above the water.

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Example�

Solve by completing the square

Check with quadratic formula

Solution

Let a = 2, b = −5, and c = −9.

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Example�

Solve by completing the square

Check using quadratic formula

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Example�

Use the discriminant to determine the number of real solutions to the quadratic equations below

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Modeling Projectile Motion�

Example 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?

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Height of a toy rocket
t (sec)	0	1	2
s(t) feet	12	36	28

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Solution continued�

If t = 0, then s(0) = 12, so

The value of v_o 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.

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Solution continued�

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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Example

A box is is being constructed by cutting 2 inch squares from the corners of a rectangular sheet of metal that is 6 inches longer than it is wide. If the box has a volume of 224 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.

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x + 10

x

x - 4

x + 6

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Solution continued�

Since the height times the width times the length must equal the volume, or 238 cubic inches, the following can be written

Step 3: Write the quadratic equation in the form ax² + bx + c = 0 and factor.

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Solution continued�

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 , which checks.

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Finding Domain Refresher

A Polynomial

All real numbers

A Rational Function

Denominator cannot be zero

An Even Radical Function

(Something) must be greater than or equal to zero

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Finding Domain

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Homework

Starting on Page 201

1, 5, 9, 13, 17, 25, 45,47, 49, 51, 53, 55, 57, 59, 61, 63, 69, 77, 79, 81, 83, 91, 93, 95, 109, 113

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Quadratic Inequalities

Solve quadratic inequalities graphically

Solve quadratic inequalities symbolically

3.3

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Quadratic Inequalities�

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.

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Graphical Solutions�

a = 1 parabola opens up

x-intercepts: −1 and 2

equation x² − x − 2 = 0

x² − x − 2 < 0,

{x|−1 < x < 2} set notation

(−1, 2) interval notation

x² − x − 2 > 0

{x|x < −1 or x > 2} set notation

(-∞,-1) U (2, ∞) interval notation

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Example�

3x² + x − 4 = 0
3x² + x − 4 < 0
3x² + x − 4 > 0

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Solution continued�

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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y < 0

y > 0

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Example�

2x² - 3x − 2 = 0
2x² - 3x − 2 < 0
2x² - 3x − 2 > 0
-3x ≥ 9 -12x²
2x² > 16
x²– 9 < 0

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Example�

The quadratic inequality

can be used to compute stopping distances in feet for a car traveling x miles per hour on dry, level pavement. Solve the inequality shown below to determine safe speeds on a curve where a driver can see the road ahead for at most 150 feet.

We locate the point of intersection

where x is positive.

This occurs when x ≈ 31.23.

Safe speeds are less than 31 mph.

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Example�

Solve x² > 7x – 10 symbolically. Write the solutions in interval notation.

Solution

Step 1: Rewrite the inequality as x² – 7x + 10 > 0

Step 2: Solve x² – 7x + 10 = 0

(x – 5)(x – 2) = 0

x = 5 or x = 2

Step 3: These two boundary numbers separate the number line into three disjoint intervals.

(–∞, 2), (2, 5), and (5, ∞)

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Solution continued�

Step 4: Choose test values.

The expression is positive when x < 2 or x > 5.

The solution set is (–∞, 2) ∪ (5, ∞).

The boundary numbers are not included because the inequality involves >.

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Interval	Test Value x	x² – 7x + 10	Positive or Negative?
(–∞, 2)	0	10	Positive
(2, 5)	3	–2	Negative
(5, ∞)	6	4	Positive

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Example�

Solve

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Homework

Starting on Page 213

1, 3, 5, 7, 11, 17, 21, 23, 27, 31, 35, 39, 45, 49

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Transformations of Graphs

Graph functions using vertical and horizontal � translations
Graph function using stretching and shrinking
Graph function using reflections
Combine transformations

3.4

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Vertical Shifts�

A graph is shifted up or down. The shape of the graph is not changed—only its position.

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Horizontal Shifts�

A graph is shifted left or right.

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Example�

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 = x² y = (x + 3)² y = (x + 3)² − 2

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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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Stretching and Shrinking�

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Horizontal Stretching and Shrinking�

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Example�

Use the graph of y = f(x) to sketch the graph of each equation.

a) y = 2f(x) b)

a) Vertical stretching

Multiply each y-coordinate

on the graph by 2.

(−2, 1 ⋅ 2) = (−2, 2)

(0, −2 ⋅ 2) = (0, −4)

(2, 1 ⋅ 2) = (2, 2)

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(−2, 1)

(0, −2)

(2, 1)

y = f(x)

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Solution continued�

b)

Horizontal stretching

Divide each x-coordinate

by ½.

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(−2, 1)

(0, −2)

(2, 1)

y = f(x)

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Reflections of Graphs�

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Example�

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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Order of transformation

horizontal shift(left/right)
horizontal stretch/compress
reflect across y-axis
vertical stretch/compress
reflect across x-axis
vertical shift(up/down)

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Combining Transformations continued�

y = −3(x + 3)² + 1

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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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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.

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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Example�

Describe how the graph of the equation

can be obtained by transforming the graph of y = |x|. Then graph the equation.

Solution

Reflect the graph across the y-axis.

Shift the graph left 3 units.

Shift the graph down 2 units.

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Given graph is y = f(x)

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Find:

y = f(x) + 2

y = f(x-2) – 1

y = -f(x)

y = f(x+1) + 2

y = 2 f(x)

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Functions you need to know!!!

Appendix A: A Library of Functions

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