# Chapter 5 Quadratic Equations and Functions

Document Sample

```					Chapter 5 Quadratic Equations and Functions

5-1 Modeling Data With Quadratic Functions

axis of symmetry           vertex

Quadratic     ax2 + bx + c; a, b, c are real #s; a  0
Function

Ex. #1, p. 234, Classifying Functions
a. y = (2x + 3)(x – 4)

b. f(x) = 3(x2 – 2x) – 3(x2 – 2)

Check Understanding:
a. f(x) = (x2 + 5x) – x2

b. f(x) = (x – 5)(3x – 1)

c. f(x) = x(x + 3)
5-1, p. 2
Parabola      graph of a quadratic function

Axis of  fold line, where each side of parabola
Symmetry is the mirror image of the other side

Vertex        turning point, where parabola crosses its
axis of symmetry

Minimum       lowest point of a parabola; when
Point         parabola opens upward

Maximum highest point of a parabola; when
Point   parabola opens downward

Ex. #2, p. 235, Points on a Parabola
Look at the graph, identify: vertex, axis of symmetry

Check Understanding; identify vertex, aos, P, Q
a.                       b.

Ex. #3, p. 237, Finding a Quadratic Model
Substitute: (2, 3); (3, 13); (4, 29) into y = ax2 + bx + c

CU: Find a quadratic function with a graph that includes (1, 0);
(2, -3); (3, -10)

Assignment: p. 237-238, 2-20, 24-36, 44-48 even
***50-58 even
5-2 Properties of Parabolas

Quadratic    y = ax2 + bx + c
Equation

Graph the following equations on graphing calculator.
1
y=x 2
y = -x 2
y = 2x 2      y  x2
2

1 2
y=-2x2        y  x            y = x2 + 3   y = x2 – 2
2

y = x2 – 2x – 3       y = -x2 + 4x + 2

Properties
a>0
a<0
b
Axis of symmetry       x
2a
 b  b 
Vertex     , f   
 2a  2a  

Y-intercept (0, c)

Assignment: p. 244-247, 2, 4, 8, 10, 14, 16, 28, 30, 36, 38, 74-
78 even
5-3 Translating Parabolas

Standard form y = ax2 + bx + c

Vertex form        y = a(x – h)2 + k

Properties
h : horizontal shift (+ right, - left)
k: vertical shift (+ up, - down)
Vertex: (h, k)

1
Ex. #1, p. 249: Graph y   ( x  2) 2  3
2

C.U. Graph y = 2(x + 1)2 – 4
5-3, p. 2
Ex. #2, p. 249 Writing the Equation of a Parabola
See graph p. 249

C. U. Write equation from graph on p. 250

Ex. #3, p. 250 Real World Connection

C. U. Write a function with the towers being 4000’ apart and
600’ high

Ex. #4, p. 251 Converting to Vertex form
Write, y = 2x2 + 10x + 7 in vertex form
Find the vertex.

C. U. Write y = -3x2 + 12x + 5 in vertex form.

Assignment: p. 251-254, 2-34, 44-56, 86-90
***70-76, 80-84 even

Vocabulary factoring          GCF
perfect square trinomial
difference of 2 squares

Factoring    rewriting an expression as the product
of its factors

Greatest     GCF; common factor with the greatest
Common       coefficient and greatest exponent
Factor       ex. 4x2 and 6x GCF = 2x

Steps for     1. Find GCF
Factoring     2. Factor into a (binomial)(binomial)
Ex. #1, p. 255 Finding Common Factors
a. 4x2 + 20x – 12     b. 9n2 – 24n

C. U. Factor each expression
a. 9x2 + 3x – 18      b. 7p2 + 21       c. 4w2 + 2w

Ex. #2, p. 256 Factoring when c > 0 and b > 0
x2 + 8x + 7

C. U. Factor
a. x2 + 6x + 8    b. x2 + 12x + 32      c. x2 + 14x + 40
5-4, p. 2
Ex. #3, p. 256 Factoring when c > 0 and b < 0
Factor x2 – 17x + 72

C. U. Factor
a. x2 – 6x + 8          b. x2 – 7x + 12    c. x2 – 11x + 24

Ex. #4, p. 257 Factoring when c < 0
Factor x2 – x – 12

C. U. Factor
a. x2 – 14x – 32        b. x2 + 3x – 10    c. x2 + 4x – 5

Ex. #5, p. 257 Factoring a ≠ 1 and c > 0
Factor 3x2 – 16x + 5

C. U. Factor
a. 2x2 + 11x + 12       b. 4x2 + 7x + 3 c. 2x2 – 7x + 6

Ex. #6, p. 258 Factoring when a ≠ 1 and c < 0
Factor 4x2 – 4x – 15

C.U. Factor
a. 2x2 + 7x – 9     b. 3x2 – 16x – 12      c. 4x2 +5x – 6
5-4, p. 3
Perfect        3 terms, 1st and last terms are perfect squares
Square         ex. 64x2 – 16xy + y2
Trinomial

Property       a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2

Ex. #7, p. 258 Factoring Perfect Square Trinomials
Factor 9x2 – 42x + 49

C. U. Factor
a. 4x2 + 12x + 9     b. 64x2 – 16x + 1        c. 25x2 + 90x + 81

Difference of      2 terms, 1st and last terms are perfect
Two Squares        squares with subtraction between them
Ex. 25s2 – t2

Property           a2 – b2 = (a + b)(a – b)

C. U. Factor
a. x2 – 64                   b. 4a2 – 49

Assignment: p. 259-261, 2-62, 80-82 even
***72-78, 86-90 even

Vocabulary standard form of quadratic equation
zero product      zero of a function

Standard form ax2 + bx + c = 0
Equation

Zero Product      If ab = 0, then a = 0 or b = 0
Property          ex. If (x + 3)(x – 7) = 0,
then (x+3) = 0 or (x – 7) = 0

To Solve          1. write equation in standard form
2. Factor
3. Set each factor = 0 and solve each
(multiple solutions written only once)
4. Check

Ex. #1, p. 263 Solve by factoring
2x2 – 11x = - 15

C. U. Solve
a. x2 + 7x = 18       b. 2x2 + 4x = 6         c. 16x2 = 8x
5-5, p. 2
Ex. #3, p. 264 Real World Connection

C. U.
a. A smoke jumper jumps from 1400’. The function describing
the height is y = -16t2 + 1400, how long with the jumper be free
falling if his parachute opens at 1000’?

Zero of a    where the graph of the function intersects
Function     the x-axis (the value of function = 0)

Ex. Graph Ex. #1 (above) 2x2 – 11x = - 15

C. U. Graph
a. x2 + 7x = 18       b. 2x2 + 4x = 6       c. 16x2 = 8x

Assign.: p. 266-268, 2-18, 36-38, 42, 46-48, 52, 68-72 even
***74-82 even
5-6 Complex Numbers

Vocabulary      imaginary #       complex #

Imaginary #     any # whose square root = -1
if i 2  1, then i   1
Ex.  4   1  4  i  2  2i
a complex #, a + bi; b  0; used when
taking square root of negative numbers
i2 = -1; i   1 ;           i   1
principal root     negative root

Ex. #1, p. 270 Simplifying Numbers Using i
Simplify  8

C.U. Simplify
a.  2              b.     12             c.    36

Complex #s      both imaginary and real #s together;
written in the form: a + bi

Name squares from 1-20:
5-6, p. 2
Ex. #2, p. 271 Simplifying Imaginary Numbers
Write  9  6 in a + bi form

C.U. Write  18  7 in a + bi form

See Diagram of Complex Numbers, p. 271

Ex. #4 Additive Inverse of a Complex Number
Find the additive inverse of -2 + 5i

C.U.
a. -5i             b. 4 – 3i            c. a + bi

Simplify: (5 + 7i) + (-2 + 6i)

C.U.
a. (8 + 3i) – (2 + 4i)         b. 7 – (3 + 2i)      c. (4 – 6i) + 3i

Assign., p. 274-276, 2-18, 24-34, 58, 60, 76 even
5-6B Operations and Absolute Value of Complex Numbers

Vocabulary           complex # plane
absolute value of a complex #

Find:         i          i2       i3       i4

i5         i6       i7       i8        Pattern?

Complex # x-axis = Real axis
Plane     y-axis = Imaginary axis (p. 271)

Absolute          a  bi  a 2  b2
Value of      Ex. |5i| =             |3 – 4i| =
Complex #
C.U. p. 271, #3 Find the absolute value:
a. |6 – 4i|             b. |-2 + 5i|            c. |4i|

Ex. #6, p. 272 Multiplying Complex Numbers
a. (5i)(-4i)              b. (2 + 3i)(-3 + 5i)

C.U. Simplify each expression:
a. (12i)(7i)    b. (6 – 5i)(4 – 3i)             c. (4 – 9i)(4 + 3i)
5-6B, p.2
Ex. #7 Finding Complex Solutions
Solve. 4x2 + 100 = 0

C.U. Solve.
a. 3x2 + 48 = 0        b. -5x2 – 150 = 0       c. 8x2 + 2 = 0

Ex. #8 Real World Connection
f(z) = z2 + i
f(0) =              first output becomes second input

f(i) =

f(-1 + i)

C.U. Find the first 3 output values for f(z) = z2 – 1 + i

Assign., p. 274-276, 20-22, 36-46, 50-52, 62-66, 78 even
***48, 68, 70, 80, 84, 86
5-7 Completing the Square

Vocabulary       completing the square

Completing       creating the left side of the equation
the Square       so that it is a square

Steps            Set equation = to constant (c); a must = 1
Take half of b and square it
Add that # to both sides of equation
Take square root of both sides (+ roots)
Check
Ex. #1, p. 278 Solving a Perfect Square Trinomial
x2 + 10x + 25 = 36

C.U. Solve: x2 – 14x + 49 = 81

Ex. #2, p. 278 Completing the Square
x2 – 8x + ___                  C.U. x2 + 7x + ___
5-7, p. 2
Ex. #3, p. 279 Solving by Completing the Square
x2 – 12x + 5 = 0

C.U.
a. x2 + 4x – 4 = 0             b. x2 – 2x – 1 = 0

Ex. #4 Finding Complex Solutions
x2 – 8x + 36 = 0

C.U.
x2 + 6x = -34

Assign., p. 281-283, 2-20, 64, 66 even
5-7B Completing the Square When a  1

Ex. #5 & 6 Solving when a  1 ; solve for y or set equation = 0,
convert to vertex form.
5x2 = 6x + 8                y = x2 + 6x + 2

C.U.
a. 2x2 + x = 6                     b. y = x2 + 5x + 3

p. 577 Example 1 and 2.
Solve.                         Graph to find solution.
x2 – y2 = 9                    2. x2 + y2 = 36
x2 + 9y2 = 169                     y = (x – 2)2 – 3
5-7B, p.2
Ex. #7 Real World Connection
p = -s2 + 120s – 2000

C.U. Using the vertex form, find the vertex
1
a. p   s 2  280s  1200           b. max or min value?
2
Why?

#41 Solve for k: x2 – kx + 100 = 0

Assign., p. 281-283, 22-36, 40-48, 52-56 even
***70-76 even

 b  b 2  4ac
2a
Formula

Steps            Write equation in ax2 + bx + c = 0 form
Substitute a, b, c values into formula
Solve

Ex. #1, p. 286 Solve 3x2 – 5x = 2

C.U. #1 Solve. 3x2 – x = 4
5-8, p. 2
Ex. #2, p. 286 Finding Complex Solutions
2x2 = -6x – 7

C.U. Solve: -2x2 = 4x + 3 and Graph
Ex. #3, p. 287 Real World Connection

C.U.
a. 4x2 = 8x – 3               b. x2 + 4x = 41

Assign., p. 289-291, 2-20, 78 even
5-8B The Discriminant

Vocabulary          discriminant

Discriminant        b2 – 4ac; part of the quadratic equation

Graph
Determine # of b – 4ac > 0; 2 Real Solutions
2

Real Solutions               2 x-intercepts
& graph in      b2 – 4ac = 0; 1 Real Solution
relation to                   1 x-intercept
x-axis         b – 4ac < 0; NO Real Solution
2

no x-intercept

Ex. #4, p. 288 Using the Discriminant
Determine the type and # of solutions: x2 +6x + 8 = 0

C.U.
a. x2 +6x + 9 = 0                     b. x2 + 6x + 10 = 0

Ex. #5 Real World Connection
h = -16t2 + 34t                       C.U. ht = 15’?

Assign., p. 289-291, 32-38, 42, 50, 58-62, 66, 76
***82-86 even

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 46 posted: 8/2/2010 language: English pages: 21
How are you planning on using Docstoc?