Transcript Ch 1 – Functions and Their Graphs

Ch 1 – Functions and Their Graphs
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Different Equations for Lines
Domain/Range and how to find them
Increasing/Decreasing/Constant
Function/Not a Function
Transformations
• Shifts
• Stretches/Shrinks
• Reflections
• Combinations of Functions
• Inverse Functions
Ch 1 – Functions and Their Graphs
1.1 Formulas for lines
y y
slope
m 2 1
x2  x1
vertical
line
xa
pointslope
y2  y1  mx2  x1 
horizontal
line
yb
slopeintercept
y  m x b
parallel
slopes
m||  m
Ax  By  C  0
perpendicular
slopes
general
form
m  
1
m
1.2 Functions
domain (input)
range (output)
3,4
1.2 Functions
domain (input)
[1, )
range (output)
[3, )
[ inclusive
( exclusive-  alway exclusive
1.2 Functions
Increasing/decreasing/constant
on x-axis only
(from left to right)
 1,  
( ) always
[ ] never
1.2 and 1.3 Functions
Functions
2,3, 4,3
Not functions
2,3, 2,4
3
2
3
4
2
4
1.2 and 1.3 Functions
Function or Not a Function?
Domain?  , 
Range?
[3, )
y-intercepts?
x-intercepts?
increasing?
decreasing?
0,1
1,0 and 5,0
2, 
 ,2
1.2 and 1.3 Functions
Finding domain from a given function.
Domain =  ,  except:
x in the denominator
3x
f x   2
x 4
Can’t divide by zero
domain:  , ; x  2 or  2
x in radical
f x   2 x  6
Can’t root negative
2x  6  0
domain: 3, 
1.4 Shifts (rigid)
y  k  ax  h
2
y  0  1x  0
2
2


y  0 1 x  2
horizontal
shift
y  2  1x  0
vertical shift
2
1.4 Stretches and Shrinks (non-rigid)
vertical
y  c x 
2
horizontal
y  3x 
stretch
2
1 2
y  x 
3
shrink
y  3x 
shrink
2
1 
y   x
3 
stretch
2
1.4 Reflections
hx  x
In the x-axis
In the y-axis
y x
hx   f x
y  x
hx  f  x 
If negative can be move to other side, flipped on x-axis.
If can’t, flipped on y-axis.
1.5 Combination of Functions
 f  g x  f g x
Give f x  x  2 and g x  3x, find  f  g  2
g  2  3 2  6
f g  2  f  6  4
1.5 Combination of Functions
 f  g x  f g x
Give f x  x  4 and g x  2 x, find g  f  2
f  2   2  4  6
g  f  2  g  6  12
1.6 Inverse Functions
Find theinverse f x   2 x  4 and verify the functions
are inversesof each other.
1. Replace f x with y and switch x and y.
x  2y  4
2. Solve for y.
x  4  2y
x4
y
2
x4
g x  
2
Show that f  g x  x
 x4
 f  g x   2
4
 2 
 f  g x  x  4  4
 f  g x  x
Ch 2 – Polynomials and Rational Functions
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Quadratic in Standard Form
Completing the Square
AOS and Vertex
Leading Coefficient Test
Zeros, Solutions, Factors and x-intercepts
Given Zeros, give polynomial function
Given Function, find zeros
Intermediate Value Theorem, IVT
Remainder Theorem
Rational Zeros Test
Descartes’s Rule
Complex Numbers
Fundamental Theorem of Algebra
Finding Asymptotes
Ch 2 – Polynomials and Rational Functions
2.1 Finding the vertex of a Quadratic Function
1. By writing in standard form (completing the square)
f x   2 x  8 x  7
2
f x  2x 2  4x  4 7  8
f x  2x  2 1
2
 2,1

2. By using the AOS formula

x
8
 2
22
b
x
2a
f x  22  82  7  8  16  7  1
2
2.1 Writing Equation of Parabola in Standard Form
Given a parabola with vertex at 1,2 and passes through
3,-6, writeits equationin standardform.
Substitute x, y, h and k intostandardformand solvefor a.
y  ax  h  k
2
 6  a3 1  2
2
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 6  a4  2
 8  4a
a  2

y  2x 1  2
2
2.2 Leading Coefficient Test
f x  ax  ...
n
Leading exponent n
Odd
Even
Leading Coefficient a
Positive
Negative
2.2 Zeros, solutions, factors, x-intercepts
There are 3 zero (or roots),
solutions, factors,
and x-intercepts.
x  2 is a zero of the function
x  2 is a solution of the function
x  2 is a factorof thefunction
2,0 is an x - interceptof thefunction
2.2 Zeros, solutions, factors, x-intercepts
Find the polynomial functions with the following
zeros (roots).
1
x   , 3, 3
2
If the above are zeros, then the factors are:
1

f x    x  x  3x  3
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2

Can be rewritten as
f x   2 x  1x  3x  3  2 x3  11x 2  12x  9
2.2 Zeros, solutions, factors, x-intercepts
Find the polynomial functions with the following
zeros (roots). x  3,2  11,2  11
Writing the zeros as factors:
 
 

f x  x  3 x  2  11 x  2  11

Simplifying.
f x  x  3 x  2  11 x  2  11

f x   x  3x  2
2

11

f x  x  3x 2  4x  4 11 x  3x 2  4x  7
f x  x3  7 x 2  5x  21
2.2 Intermediate Value Theorem (IVT)
IVT states that when y goes from positive to negative,
There must be an x-intercept.
2.3 Using Division to find factors
Long Division
Synthetic Division
2.3 Remainder Theorem
When f x is divided by x  k  then f k  is theremainder.
Is x  2 a factorof f x  3x3  8x 2  5x  7 ?
Using syntheticdividion : - 2
3
8 5 -7
-6 -4 -2
3 2 1 -9
 9 is theremainder
T herefore,x  2 is not a factor.
Also, - 2, - 9 must be a pointon thegraph.
2.3 Rational Zeros Test
f x  qxn  ... p
p
factorsof constant erm
t
Possible RationalZeros 
q factorsof leading coefficient
Find thepossible rationalzerosof f x  2x3  3x 2  8  3.
Factorsof 3  1,3
1 3

 1,3, ,
Factorsof 2  1,2
2 2
2.3 Descartes’s Rule
Count number of sign changes of f(–x) for number of
positive zeros
f x  3x3  5x 2  6 x  4
+
–
+ –
1
2
3 = 3 or 1 positive zeros
Count number of sign changes of f(–x) for number
of negative zeros.
f  x  3 x  5 x  6 x  4
3
2
–
–
–
0 negative zeros
–
(+) (–) (i)
3
0
0 3
1
0
2 3
2.3 Complex Numbers
Complex number = Real number + imaginary number
Multiply3  5i by its conjugate.
3  5i 3  5i 
Treat as difference of squares.
32  5i2  9   25  34
2.3 Complex Numbers
3  5i
Write in standard form
.
4  2i
3  5i 4  2i 12  6i  20i  10 2  26i 1 13



  i
4  2i 4  2i
16   4
20
10 10
2.5 Fundamental Theorem of Algebra
A polynomial of nth degree has exactly n zeros.
f x   5 x 4  x 3
has exactly 4 zeros.
2.5 Finding all zeros
f x  x5  x3  2x 2 12x  8
1. Start with Descartes’s Rule
2. Rational Zeros Test (p/q)
+
–
i
2
1
2
0
1
4
 1,2,4,8
PRZ 
 1,2,4,8
1
3. Test a PRZ (or look at graph on calculator).
x  1
1
x  1
1
1 0 1
1 1
1 1 2
1 2
1 2 4
2  12 8
2 4 8
4 8 0
4 8
8 0
x  2
2
1 2 4 8
 2 08
1 0 4 0
x2  4
x  2i,2i
2.6 Finding Asymptotes
N axn  ...
f x    m
D bx  ...
Vertical Asymptotes
Where f is undefined. Set denominator = 0
Horizontal Asymptotes
Degree larger in D, y = 0. BOBO n  m
Degree larger in N, no h asymptotes. BOTN n  m
Degrees same in N and D, take ratio of coefficients. n  m
a
y
b
Ch 3 – Exponential and Log Functions
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Exponential Functions
Logarithmic Functions
Graphs (transformations)
Compound Interest (by period/continuous)
Log Notation
Change of Base
Expanding/Condensing Log Expressions
Solving Log Equations
Extraneous Solutions
Ch 3 – Exponential and Log Functions
3.1 Exponential Functions
Same transformation as hx   k  f x  h
If negative can be move to other side, flipped on x-axis.
If can’t, flipped on y-axis.
f x   3x1  2 Shifted 1 to right, 2 down.
f x   3x Flipped on x-axis.
f x   3
x
Flipped on y-axis.
3.1 Compounded Interest
Compound by Period
 r
A  P1  
 n
nt
Compound Continuously
A  Pert
3.1 Compounded Interest
A total of $12,000 is invested at an annual interest rate
of 3%. Find the balance after 5 years if the interest is
compounded (a) quarterly and (b) continuously.
nt
 r
A  P1  
P  12,000
 n
r  0.03 per year
 0.034 5
n  4 timesper year A  12,0001
  13,934.21
4 

t  5 years
A  Pert

A  12,000e0.035  13,942.01
3.2 Logarithms
Used to solve exponential problems (when x is an
exponent).
x  ay
y  loga x
3.2 Logarithms
Used to solve exponential problems (when x is an
exponent).
xa
y
y  loga x
Change of base
logb
loga b 
loga
3.3 Logarithms
Expanding Log Expressions
3
5x
log4
 log4 5  3 log4  log4 y
y
Condensing Log Expressions

x  2
2 lnx  2  ln x  ln
2
x
3.4 Solving Logarithmic Equations
Solve the Log Equation
 
3 2  42
x
x in the exponent, use logs
2 x  14
ln 2  ln 14
x
x ln 2  ln 14
ln 2 ln 14
x

ln 2 ln 2
x  3.807
3.4 Solving Logarithmic Equations
Solve the Log Equation
e  3e  2  0
2x
e
x
x


 2 e 1  0
e
x
x

2  0
e
x

1  0
ex  2
e 1
ln e  ln 2
ln e x  ln 1
x
x  ln 2
x
x  ln 1  0
3.4 Solving Logarithmic Equations
Solve the Log Equation
2 log5 3x  4
log5 3x  2
5
log5 3 x
5
3x  25
25
x
3
2
3.4 Solving Logarithmic Equations
Solve the Log Equation
lnx  2  ln2 x  3  2 ln x
lnx  22 x  3  ln x 2
x  22x  3  x2
2x  7 x  6  x
2
x  6,1
ln1  2  invalid
2
x  7x  6  0
2
x  6x 1  0
x6
3.4 Solving Logarithmic Equations
Solve the Log Equation
lnx  2  ln2 x  3  2 ln x
lnx  22 x  3  ln x 2
x  22x  3  x2
2x  7 x  6  x
2
x  6,1
ln1  2  invalid
2
x  7x  6  0
2
x  6x 1  0
x6