Download 2nd Quarter Notes

2nd Quarter Notes
11/10/08
Review Quadratic Equation
X²-8X+15 = 0
Standard Form: ax²+ bx + c
Four ways to solve the quadratic equation
1. Graph it
2. Factor it.
3. Use the quadratic formula X =
 b  b² - 4ac
2a
4. Completing the square.
1) Factoring Method: X² -8X + 15
(X-3)(X-5)
X=3
X=5
2.) Quadratic formula
 b  b² - 4ac
X=
2a
 (8)   8² - 4(1 * 15)
X=
2(1)
X=
X=
b²- 4ac > 0
2 real solutions
b²- 4ac < 0
1 solution
b²- 4ac = 0
no real solution.
8 4 82
8  64  60

=
2
2
2
10
5
2
X=
6
3
2
Example 1: 2X² + 5X – 3
Factor What numbers multiplied =3 and added = 5
(2X -1)(X +3) The larger number must be – because middle value is negative.
2X – 1 =0
+1 +1
2X = 1
2
2
X = .5
X+3=0
- 3 -3
X = -3
3.) Completing the square
2X² - 6X + 1 = 0
-1 -1
Subtract 1
2X² - 6X = -1
2
2
Divide by 2 to get rid of the 2 in 2X²
3
9
b
Use   ² ²  = 2.25 add to both sides.
2
4
2
3
X²- 3X + 2.25 = 2.25 - .5 Use
for factors.
2
3
(X - ) ² = 1.75
2
X² - 3X = -.5
3
 1.75
2
X = 1.5 + 1.32 = 2.82
X=
X = 1.5 – 1.32 = .22
Example 2: Completing the square
4X²- 13X + 3 = 0
- 3 -3
4X²- 13X = -3
4
4
Subtract 3 from both sides.
Divide both side s by 4
  3.25 
Use (b/2)² 
 ² = 1.625 add to both
 2 
sides.
X²- 3.25X + 1.625 = 1.625 - .75
(X – 1.625)² =  .875
X – 1.625 =  .935
X = 1.625 + .935 = 2.56
X = 1.625 - .935 = .69
X² - 3.25X
= -.75
Quadratic Formula
X=
 b  b² - 4ac
2a
2X² - X + 5
a = 2 b = -1
c=5
Replace values of a,b,c in quadratic formula.
X=
 (1)  (1)² - 4(2)(5)
2(2)
X=
1  1  40
4
X=
1   39 1  6.29i

4
4
X=
7.29i
= 1.82i
4
Discriminant
X=
a=2, b=-1, c=5
5.29i
 1.32i
4
b² - 4ac
No real solutions
b² - 4ac > o 2 real solutions
b² - 4ac = 0 1 solution
b² - 4ac < 0 no real solutions
11/13/08
Domain Range
X
Y
2
4
3
8
4
8
5
16
6
21
This is a function
Which one is a function?
4X – 2Y² + 5 = 0
- 4X
-5
-4X
-2Y²
= - 4X – 5
-2
-2 -2
Y²= 2X + 2.5
Y=
Input Output
X
Y
2
3
3
9
3
12
4
15
5
18
Not a function – the X value repeats.
Put into Y = form.
Y 2  2 X  2.5 To see if a function give a value for X.
Y = 2 X  2.5
2(9)  2.5  18  2.5  20.5
Y =  4.52
Example 2:
4X – 2X³+ 5 = 0
-4X
- 5 -5 - 4X
-2X³
= -4X -5
-2
-2 -2
X has 2 values so this is not a function.
3
X³= 2X – 2.5
Y  3 2 X  2.5
Y=
3
2(4)  2.5  3 8  2.5  3 10.5
Y = 2.18
One answer, this is a function.
11/14/08
Quadratic Function graph
Graph is a parabola.
Top graph – has a minimum.
Vertex is (0,0) also y-intercept.
Bottom graph is a maximum
Vertex is (1.5, 0) X- intercept is 1.5.
Standard Form for Quadratic Function.
aX²+ bX + c
Transformations – a,h,k,t,and s are constants (numbers).
Transformations Polynomials
x-intercept
Forms
a(X –h)²+k
ax²+bx + c
a(x-s)(x-t)
Vertex
(h,k)
b
b
,f(
)
2ac
2ac
st
st
,f(
)
2
2
 b  b 2  4ac
2a
s and t
x-intercept
h
y-interept
k
a
ah²+ k
C
Transformation
f(X) = 2(X-3)²-4
a = 2 h = 3 k = -4
Vertex: (3,-4)
x-intercept: h 
k
 (4)
 3
 3 2
a
2
3+ 1.41 = 4.41
3 – 1.41 = 1.59
y-intercept: ah²+k = 2(3²) + -4 = 18 – 4 = 14
a●s●t
Polynomial Form
f(x) = X²+ 2X +3
Vertex:
a = 1 b = 2 c =3
b
2 2


 1
2ac 2(1)
2
f(-1) = (-1²)+2(-1)+3
f(-1)= 1 - 2+3= 2
Vertex: (-1,2)
X – intercept:
 b  b 2  4ac  2  22  4(1)(3)  2  4  12  2   8
=


2(1)
2
2
2a
Can’t take √ of a negative number, no X-intercept.
Y-intercept: C = 3
X-Intercept
f(X)= a(x-s)(x-t)
f(X) =
1
( X  4)( X  2)
2
Vertex:
42 2
 1
2
2
Vertex (1, 4.5)
a=
1
2
s=4
t = -2
f(1) = -.5(1-4)(1+2)= -.5(-3)(3)= 4.5
X-intercept: (s and t) 4 and -2
Y-intercept: a(s)(t) = -.5(4)(-2)= 4
Changing Forms
f(X)= .4(X -3)² +2
= .4 (X -3)(X -3) + 2
= .4 (X² -3X – 3X + 9) +2
= .4 (X²-6X +9) +2
= .4 X²-2.4X +3.6 +2
= .4 X²-2.4X + 5.6
f(X) = 3 X² -3.9X – 43.2
transformation
now a polynomial
Polynomial
=
3X² 3.9 X 43.2


3
3
3
Factor out the 3
= 3(X²-1.3X – 14.4)
= 3(X-4.5)(X+3.2) X-intercept form
X-intercept Form
Polynomial
h(X) = -2 (X -4)(X +2)
= -2(X² -4X +2X -8)
= -2(X²-2X-8) = 2 X²+ 4X +16
f(X) = (-3X² + 4X) -1 Polynomial – Complete the square.
= -3(X² - 1.33X +
Factor out the 3
b
1.33 2
(
)  (.67 2 )  .44
2
2
=-3(X² - 1.33X + .44 - .44) – 1
= -3(X-.67)(X - .67) – 1
= -3(X-.67)²- 1
Transformation
11/18/08
Concave down
Concave up
Parametric Equation – comes in twos/ with t
Y = f(X)
X=t
Y = f(t)
X =2t +1 Y = t²-3
Calculator: Mode → Function → Par enter
Y= put in equation
X = 2t +1
Y = t²-3
Go back to Mode →Function
-2X³+6X²-X + 3 3 is Y – intercept
Points of inflection – where the shape starts to change.
How to graph without a calculator.
X = 2t +1
Y = t²- 3
Make a table:
14
T X=2t +1
Y=t²- 3
1
Y = 1²-3=-2 3,-2
X=2(1)+1=3
X,Y
2
X=2(2)+1=5 Y=2²-3=1
5,1
3
X=2(3)+1=7 Y=3²-3=6
7,6
4
X=2(4)+1=9 Y=4²-3=13
9,13
12
10
8
6
Series1
4
2
0
-2 0
5
10
-4
Pg. (10-14)(24-28)(62-67)
11/21/08
Solve the equations
Ex. 1 5X²+7X-10 =0
+10 +10
5X²+7X =10
5
5
X²+ 1.4X
=2
Complete the square take b/2
X²+1.4X+.49=2+.49
(X+.7)²= 2.49
X+.7 =  2.49
X+.7 = 1.57
X+.7 = -1.57
-.7 -.7
-.7 -.7
X = .87
X = -2.27
(1.4) 2
 .7 2  .49
2
Example 2:
 b  b 2  4ac
X=
2a
12X²+6X+13=0
a=12, b=6, c=13
 6  62  4(12)(13)  6  36  624  6   588
X=


2(12)
24
24
 6  24.24
- 30.24i
 1.26 i
i =
24
24
NO Real Solution
X=
X=
18.24i
 .76 i
24
Operation of Functions
Addition
f(x) = 4X³+3X²+X
f(x) + g(x) = h(x)
g(x) = 2X²+3X-6
4X³+3X²+X
2X²+3X-6
h(x) 4X³+5X²+ 4X -6
h(x) = 4X³+5X²+ 4X -6
Subtraction
l(x) = f(x) – g(x)
4X³+3X²+X
(2X²+3X-6)
4X³+3X²+X
- 2X²-3X+6
l(x) = 4X³+X²-2X+6
m(x) = g(x) – f(x)
2X²+3X-6
- (4X³+3X²+X)
- 4X³-X²+2X – 6
Multiplication
f(x)= X²+6
g(x) = 3X²+X-3
l(x) = f(x) ● g(x)
(X²+6)( 3X²+X-3) = X²(3X²)+ X²(X)+ X²(-3)+6(3X²)+6(X)+6(-3)
= 3X4+X³-3X²+18X²+6X-18
l(x) = 3X4+X³+15X²+6X-18
Division
f ( x)
X 2  6 1X  3 1
f(x) = X²+6 l(x) =

 X 3
g ( x)
3X  2
3
3
g(x) = 3X+2
Composite Function
g ● f(x) = g(f(x))
f(x).
f(x) = 4X² + 1
g(x) =
Give a value for X then solve for
1
X 2
1st Take f(x) and use X=3
f(3) = 4(3²)+1 = 4(9) +1 = 37
g(x) =
1
X 2
g(37) =
Now replace X in g(x) with 37.
1
1

 .025
37  2 39
Division & Synthetic Division
Divide
3 X 4  8 X 2  11X  1 by X-2
3X³+6X²+ 4X-3
X-2 3 X 4  0 X 3  8 X 2  11X  1
Step1: Divide
Step2: Multiply(3X³)(X-2)
-3X4 –(-6X³)
Step3: Subtract
6X³-8X²
Step4: Compare & bring down
- 6X³-(-12X²)
Begin again
4X² -11X
-4X²-(-8X)
-3X +1
- (-3X)-(6)
-5 remainder
Synthetic Division: Only with linear functions.
3 X 4  0 X 3  8 X 2  11X  1 Dividend
X-2 Divisor
Answer – Quotient
Take constant from divisor, 2 and the numbers from the dividend.
X³ X² X
2
3 0 -8 -11 1
Multiply first number by the constant.
3
6
6
x2
12
8 -6
4 -3 -5 remainder Answer 3X³+6X²+4X-3 r-5
x2 x2
Example 2:
Divide X 5  5 X 4  6 X 3  X 2  4 X  29 by X+3
X4
1
-3
X³ X²
5
6
-3 -6
2
0
1
X
-1 4
0 3
-1 7
-29
21
-8
Answer X 4  2 X 3  X  7
To check your answer:
The Division Algorithm
Dividend = Divisor ● Quotient + Remainder
f(x) = h(x) ● g(x) + rx
( X 4  2 X 3  X  7 )(X+3) + 8 = X(X4)+X(2X³)+X(-X)+X(7)
+ 3(X4)+3(2X³)+3(-X)+3(7)
X5+5X4+6X³-X²+4X+29
Remainders & Factors
When the remainder is 0 the divisor and quotient are factors of the
dividend.
Divide
6X³ - 4X² +3X -2 by 2X²+1
3X
-2
2X²+1 6X³ - 4X² +3X -2
- 6X³
- +3X
-4X²
-2
4X²
+2
0
0
Answer: (2X²+1)(3X-2)
no remainder
The Remainder Theorem: If a polynomial f(x) is divided by X-C, then
the remainder is f(C) constant.
Divide f(x)=
X 79  3 X 24  5 by X -1 Constant 1
179  3(124 )  5  1  3  5  9 9 is the remainder.
Factor Theorem: A polynomial f(x) has a linear factor X-a if and only
if, f(a) = 0.
X-3 is a factor of X³- 4X² +2X +3
a=3
3³- 4(3²) + 2(3) +3 = 27 -36 +6 +3 = 0
3]
1
1
-4
2
3
3 -3
-3
-1
-1
0
1X²- 1X -1
(X-3)( 1X²- 1X -1)