1 of 18 pre-cal chapter 2 section 1 sat/act warm - up

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1 of 18

Pre-Cal Chapter 2 Section 1

SAT/ACT Warm - up

2 of 18

Pre-Cal Chapter 2 Section 1

Chapter 2 – Vocabulary Words

1. Polynomial function2. Linear function3. Quadratic function4. Continuous5. Intermediate Value Theorem6. Synthetic division7. Imaginary number8. Complex conjugates9. Rational function10. Slant Asymptote

See Page 172

Graphing Quadratic Functions

Chapter 2 Sec 1

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Pre-Cal Chapter 2 Section 1

Essential Question

How do you sketch graphs and write

equations of parabolas?

Key Vocabulary:Polynomial

FunctionLinear equationQuadratic function

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Pre-Cal Chapter 2 Section 1

Definitions

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Pre-Cal Chapter 2 Section 1

Quadratic function

The graph of a quadratic function is a parabola.Every parabola is symmetrical about a line called the axis (of symmetry).

The intersection point of the parabola and the axis is called the vertex of the parabola.

x

y

axis

f (x) = ax2 + bx + cvertex

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Pre-Cal Chapter 2 Section 1

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Pre-Cal Chapter 2 Section 1

Leading Coefficient

The leading coefficient of ax2 + bx + c is a.

When the leading coefficient is positive, the parabola opens upward and the vertex is a minimum.

When the leading coefficient is negative, the parabola opens downward and the vertex is a maximum.

x

y

f(x) = ax2 + bx + ca > 0 opens upward

vertex minimum

x

y

f(x) = ax2 + bx + c

a < 0 opens downward

vertex maximum

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Pre-Cal Chapter 2 Section 1

5

y

x

5-5

Simple Quadratic Functions

The simplest quadratic functions are of the form f (x) = ax2 (a 0) These are most easily graphed by comparing them with the graph of y = x2.

222 2 and 2

1 , xxgxxfxy

2

2

1xxf

22xxg

Example: Compare the graphs of

2xy

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Pre-Cal Chapter 2 Section 1

Example: f(x) = (x –3)2 + 2

Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis.

f (x) = (x – 3)2 + 2 is the same shape as the graph ofg (x) = (x – 3)2 shifted upwards two units. g (x) = (x – 3)2 is the same shape as y = x2 shifted to the right three units.

f (x) = (x – 3)2 + 2

g (x) = (x – 3)2y = x 2

- 4x

y

4

4

vertex (3, 2)

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Pre-Cal Chapter 2 Section 1

Completing the Square

1. Group together the x’s

2. Factor out the a.

3. Take coefficient of x.

a. Divide by 2

b. Then square result

4. Add the “squared result” inside the parenthesis and subtract the “Squared result times a” outside the parenthesis.

5. Factor the parenthesis, it will be (x ± (3a.))2.

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Pre-Cal Chapter 2 Section 1

x

y

Quadratic Function in Standard Form

Example: Graph the parabola f (x) = 2x2 + 8x + 5 and find the axis and vertex.

f (x) = 2x2 + 8x + 5 original equation

f (x) = 2( x2 + 4x) + 5 group the x’s & factor out 2

f (x) = 2( x2 + 4x + 4) + 5 – 8 complete the square

f (x) = 2( x + 2)2 – 3 standard form

a > 0 parabola opens upward like y = 2x2.

h = –2, k = –3 axis x = –2, vertex (–2, –3). x = –2

f (x) = 2x2 + 8x + 5

(–2, –3)

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Pre-Cal Chapter 2 Section 1

x

y

4

4

Vertex and x-Intercepts

Example: Graph and find the vertex and x-intercepts of f (x) = –x2 + 6x + 7.

f (x) = – x2 + 6x + 7 original equation

f (x) = – ( x2 – 6x) + 7 factor out –1

f (x) = – ( x2 – 6x + 9) + 7 + 9 complete the square

f (x) = – ( x – 3)2 + 16 standard form

a < 0 parabola opens downward.

h = 3, k = 16 axis x = 3, vertex (3, 16).

Find the x-intercepts by solving–x2 + 6x + 7 = 0.

(–x + 7 )( x + 1) = 0 factor

x = 7, x = –1 x-intercepts (7, 0), (–1, 0)

x = 3f(x) = –x2 + 6x + 7

(7, 0)(–1, 0)

(3, 16)

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Pre-Cal Chapter 2 Section 1

y

x

f (x) = a(x – h)2 + k standard form

f (x) = a(x – 2)2 + (–1) vertex (2, –1) = (h, k)

y = f(x)

(0, 1)

(2, –1)

Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1

1 = 4a –1 and 2

1a

122

112

2

1 22 xxxfxxf

Example: Find an equation for the parabola with vertex (2, –1) passing through the point (0, 1).

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Pre-Cal Chapter 2 Section 1

Example: Find the vertex of the graph of f (x) = x2 – 10x + 22.

f (x) = x2 – 10x + 22 original equation

a = 1, b = –10, c = 22

At the vertex,

So, the vertex is (5, -3).

a

bf

a

b

2,

2 is

The vertex of the graph of f (x) = ax2 + bx + c (a 0)

512

10

2

a

bx

Vertex of a Parabola

322510552

2

f

a

bf

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Pre-Cal Chapter 2 Section 1

Minimum and Maximum Values of Quadratic Functions

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Pre-Cal Chapter 2 Section 1

The path is a parabola opening downward. The maximum height occurs at the vertex.

At the vertex,

So, the vertex is (9, 15).

The maximum height of the ball is 15 feet.

629

1 2 xxy

2,9

162

9

1 2

baxxy

.9

91

2

2

2

a

bx

1592

f

a

bf

Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is:

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Pre-Cal Chapter 2 Section 1

Essential Question

How do you sketch graphs and write

equations of parabolas?

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Pre-Cal Chapter 2 Section 1

Daily Assignment

• Chapter 2 Section 1/ Ch 1 Review• Text Book (TB)

• Textbook Pgs 99– 102 #1 – 4, 7, 11, 17, 21,

27, 29, 31, 35, 55, 63 – 67 All

• Read Sections 2.2 – 2.3

• Show all work for credit.