Transcript y - GCC

x coordinates
y coordinates
Compare all the x coordinates, repeats. The set is not a function,
just a relation.
Compare all the x coordinates, no repeats. The set is a function.
Compare all the x coordinates in the
domain, only one corresponding arrow
on each x coordinate.
The set is a function.
Compare all the x coordinates in the
domain, 8 has two corresponding
arrows. Repeats
The set is not a function,
just a relation.
When determining if a graph is a function,
we will use the Vertical Line Test.
Use your pencil as a Vertical Line and place
it at the left side of the graph.
Slide the pencil to the right and see if it
touches the graph ONLY ONCE. If it does
it is a FUNCTION.
FUNCTION.
Use your pencil as a Vertical Line and place
it at the left side of the graph.
The Vertical Line crosses the graph in 2 or
more locations, therefore this graph is just a
RELATION.
Not multiplication!
y coordinates
input
output
y = f(x)
Independent Variable
Dependent Variable
y = 3(4) + 7
y = 12 + 7
y = 19
The work is the same!
f(4) = 3(4) + 7
f(4) = 12 + 7
f(4) = 19
Put ( )’s around every x.
g x   x   5x   3
g  6   6  5 6  3
2
g  6  36  30  3
g  6  3
Substitute -610
for every x.
Simplify by
Order of
Operations.
g x   x   5x   3
2
g 3t   3t   53t   3
2
g 3t   9t  15t  3
2
g x   x   5x   3
2
2
g 10  10  510  3
2
g 10  100  50  3
g 10  147
2




g x  x  5x   3
g b  2  b  2  5b  2  3
2
FOIL and distribute
b  2b  2
g b  2  b 2  4b  4  5b  10  3
2
b
 2b Like
2b Terms,
 4 CLT.
Combine
2
2


g
b

2

b
b  4b  4  9b  11
Remember h(x) = y
???
h(x) = y
h(3) = 2
=y
Find the point
when x = 3
j(3) = 5
h(x) = y
h(2) = 1
h(x) = y
h(0) is not possible! Zero is not in the Domain.
h(0) = undefined
=y
Find the point
when x = -2
j(-2) = 1
(3, 5)
=y
Find the point
when x = 0
j(0) = -1
(-2, 1)
0
-2
3
(0, -1)
Every x coordinate
from 3 to 6
(3, 5)
=y
Find the point
when y = 3
x=2
(6, 5)
5
Find the point
when y = 1
3
(2, 3)
(-4, 1)
1
x = -4, -2 & 1
(-2, 1)
(1, 1)
-3
(-?, -3)
Find the point
when y = 5
Find the point
when y = -3
3<x<6
j(x) = -3 is not possible! -3 is not in the Range.
[3, 6] interval notation
(?, 3)
Domain
Find the smallest x coordinate to the
largest x coordinate.
7
5
-7
Domain: -7 < x < 6 or [-7, 6]
3
6
Range
Find the smallest y coordinate to the
largest y coordinate.
The first set of y coordinates are
-4
-4 < y < 3 or (-4, 3). Notice that we
started and ended at open circles.
The second set of y coordinates are
5 < y < 7 or [5, 7]
Open circles mean that the point doesn’t
exist and the closed circle means that the
point is there. x = -3 at this location…as
long as we can touch the graph the x
coordinates are there and continuous.
Range: -4 < y < 3 or 5 < y < 7
(-4, 3) U [5, 7]


Domain
Find the smallest x coordinate to the
largest x coordinate.
Domain: x > -4 or [-4, oo)
Range
Find the smallest y coordinate to the
largest y coordinate.
-4
Range: y > -7 or [-7, oo)
-7
Domain
Find the smallest x coordinate to the
largest x coordinate.
4
Domain: -8 < x < 8 or [-8, 8]
1
-8
-1
8
Range
Find the smallest y coordinate to the
largest y coordinate.
The y coordinates are not connected
or consistent, therefore we list them
separately.
Range: {-1, 1, 4}
When given the function in set notation, list the x and y coordinates separately.
Domain: {-1, 1, 2, 3, 4, 5}
Range: {1, 2, 3, 4, 7, 8}
Find the domain of the functions.
When finding the domain of functions in equation form we will ask ourselves the
following questions….
Will the function work when the x is a negative?, …. a zero?, … a positive?
If the answers are 3 yes’s, then the domain is all real numbers.
If there is a no, then there is a domain restriction we need to find.
f x  4x  2 Can I multiply 4 by a negative?, a zero?, a positive?
… and then add 2 to the product?
ALL Yes!
Domain is ALL REAL NUMBERS
g x   x 2  2
h x   x 4  x 3  1
 , 
Can I square a negative?, a zero?, a positive? … and
then add 2 to the value?
ALL Yes!
Domain is ALL REAL NUMBERS
 , 
If I square a negative?, a zero?, a positive? … I should
be able to raise them to any power!
ALL Yes!
Domain is ALL REAL NUMBERS
 , 
Find the domain of the functions.
x2
j x  
x 1
k x   x  9
Adding and subtracting always is a Yes…Can I divide by a
negative?, a zero?, a positive?
NO! Can’t divide by ZERO! Set the denominator equal to zero
and solve for x to find the restriction.
Domain is ALL REAL NUMBERS, except 1
x 1
Can I take the absolute value of a negative?, a zero?, a
positive?
ALL Yes! Domain is ALL REAL NUMBERS
 , 
5
m x   2
x 9
We have a fraction again, set the bottom equal to zero
and solve for x.
x2  9  0
x  3x  3  0
x  3
Domain is ALL REAL NUMBERS, except for -3 and 3.
 ,3   3,3  3, 
8
qx   2
x  6 x  16
We have a fraction again, set the bottom equal to zero
and solve for x by factoring.
x 2  6 x  16  0
x  8x  2  0
x  8,2
Domain is ALL REAL NUMBERS, except for -8 and 2.
 ,8   8,2  2, 
2
p x   3
x  x 2  12 x
We have a fraction again, set the bottom equal to zero
and solve for x by factoring.
3
2
x  x  12 x  0

x x 2  x  12

xx  4x  3  0
x  0,3,4
Domain is ALL REAL NUMBERS, except for -3, 0 and 4.
 ,3   3,0  0,4  4, 
Domain Restrictions
5 < -5, FALSE
-5 < 5 < 3 , FALSE
5 > 3, TRUE
x = 5, test it in the
domain restrictions
to see which one is
true! Substitute
the 5 into that
function.
f 5  5  9   4
f 5  4
x = -7, and -7 < -5.
Substitute -7 into
the first function.
f  7   3 7   2
f  7   23
x = -5, and -5 < -5 < 3.
Substitute -5 into the
second function.
f  5   5  1
2
f  5  24
x = 3, and 3 > 3.
Substitute 3 into the
third function.
f 3  3  9   6
f 3  6
y  f ( x)  c
f ( x)  3
y  f ( x)  x 2
y  f ( x)  x
y  f ( x)  x1
y  f ( x)  mx  b
y  f ( x)  x n  x n1      c
y  f ( x)  x 3
Cubic Func.
y  f ( x) 
1
x
(0, 6)
rise
m = slope = run
b = y-int = (0, b)
starting point
y-int = (0, 6)
directions
-5
m=
2
down 5
right 2
down 5
right 2
up 1
right 3
right 3 up 1
point = (x1, y1)
m = slope = rise
run
starting point
y-int = (-3, 4)
directions
1
m=
3
Or in reverse
left 3
up 1
down 1
(-3, 4)
right 3
A, B, and C are integers.
To graph find x and y intercepts
???
To find the y intercept the
x coordinate is zero!
(0, y)
20  3 y  18
 3 y  18
y  6
0,6
To find the x intercept the
y coordinate is zero!
( x, 0)
2x  30  18
2x  18
x 9
9,0
Doesn’t fit, but that is
ok…we can use the slope!
 A 2 2
m


B
3 3
Notice that there is no y
variable in the equation. This
means we can’t cross the y
axis! Must be a VERTICAL
LINE at x = 6
rise
m = slope = 0 = undefined
Notice that there is no x variable
in the equation. This means we
can’t cross the x axis! Must be a
HORIZONTAL LINE at y = - 4
0
m = slope = run = 0
To graph find x and y intercepts.
We can see that 3 will divide into
-9 evenly, but 5 won’t. So we
should find the x intercept and
the slope to graph this line.
To find the x intercept the
y coordinate is zero!
( x, 0)
3x  50  9
3x  9
x  3
Find the slope!
 A 3
m

B
5
 3,0
Write the equation of a line that contains the points ( 3, 8 ) and ( 5, -1 ).
Find the slope first. m  y2  y1  1  8
9


x2  x1
53
2
Next use the point-slope form to write the equation. y  y1  mx  x1 
9
9
y  8   x  3 Convert to y = mx + b. y   x  3  8
2
2
9
27
9
43
y  x
8   x
2
2
2
2
Yellow TAXI Cab Co. charges a $10 pick-up fee and charges $1.25 for each mile.
Write a cost function, C(m) that is dependent on the miles, m, driven.
Remember…functions are equal to y. y = C(m). Use y = mx + b.
The slope is the same as rate! The y intercept (b) is the starting point or initial cost.
The $10 pick-up fee is a one time charge or initial cost. b = 10
The $1.25 for each mile is a rate. m = 1.25
Replace y with C(m) and x with m.
y  1.25 x  10
Cm  1.25m  10
In the year 2000, the life expectancy of females was 83.5. In 2004, it was 86.5.
Write a linear function E(t) where t is the number of years after 2000 and E(t) is
the life expectancy in t years. Estimate the life expectancy in the year 2009.
Estimate the when the life expectancy will be 94.
Looks difficult only because of all the words! Understand the data given to
write the equation of a line!
This looks like points (x, y) = (t, E(t))
Year # of years after 2000 (t) Age E(t)
(0, 83.5)
2000
0
83.5
2004
4
86.5
(4, 86.5)
We are back to the first problem we did for writing the equation of a line.
Use y = mx + b because we are working with functions and (0, 83.5) is the
y intercept….b is 83.5.
86.5  83.5 3
Find the slope between the points. m 
  0.75
40
4
y  0.75 x  83.5 Estimate the life expectancy in the year 2009. t  9
Et   0.75t  83.5 E9  0.759  83.5  90.25
Estimate the when the life expectancy will be 94. Et   94
94  0.75t  83.5
10.5  0.75t
t  14
 83.5 
 83.5
0.75 0.75
14 years past the year 2000, 2014.
In the year 2003, a certain college had 3450 students. In the year 2008, the
college had 4100 students. Write a linear function P(t) where t is the number of
years after 2000 and P(t) is the population of the college. Estimate the
population in the year 2012. Estimate the year when the population will reach
5400.
Understand the data given to write the equation of a line!
Year
2003
2008
# of years after 2000 (t)
3
8
Students P(t)
3450
4100
Points (x, y) = (t, P(t))
(3, 3450)
(8, 4100)
Use y = mx + b because we are working with functions, but this time we
will have to solve for b.
4100  3450 650

 130
Find the slope between the points. m 
83
5
y  130 x  b Plug in a point, (8, 4100).
Estimate the population in the year 2012.
4100  1308  b y  130 x  3060
t  12 P12  13012  3060  4620
4100 1040  b
Pt   130t  3060
3060  b
Estimate the year when the population will reach 5400.
Pt   5400
5400  130t  3060
18  t
2340  130t
18 years past the year 2000 is the year 2018.
Same Line
 4  7  3, true
5 4   7  27
 20  7  27, true
Yes, (-4, 7) is a solution.
Find the solution to the system by graphing.
x  y  1  1
m
1
1
x y 3
1
m
 1
1
( 1, 2 )
Solution is ___________.
Find the solution to the system by graphing.
y  3x  2
Convert to y = mx + b
6 x  2 y  12
 6x
 6x
2 y  6 x  12
2
2
2
y  3x  6
The slopes are the same
and the y-intercepts are
different.
Solution is ___________.
No Solution
Find the solution to the system by graphing.
2 x  3 y  6
Divide everything by 4.
8 x  12 y  24
4
4
4
2 x  3 y  6
Same LINE! Infinite Solutions,
but not the final answer.
Convert to y = mx + b
 3 y  2 x  6
2
y  x2
3
 2

 x, x  2 
Solution is ___________.
 3

Answer must be written as a point.
Section 8.2. Solving linear systems by SUBSTITUTION & ELIMINATION.
Solve for x & y.
3x  2 y  11
x y 3
Substitution Method.
1. Choose an equation and get x or y Step 1
by itself.
2. Substitute step 1 equation into the
second equation.
x y 3
y  x3
Step 2
3x  2 y  11
3 x  2 x  3  11
Step 3 3 x  2 x  6  11
3. Solve for the remaining variable.
4 Substitute this answer into the step 1 Step 4 y  x  3
equation.
y  1 3
y4
5 x  6  11
5x  5
x 1
(1,4) Is the intersection point
and solution.
Solve the system for x and y.
x  3y  1
y  2x  3
x  2y 5
2 x  4 y  11
x  32x  3  1
22 y  5  4 y  11
x  6x  9  1
4 y  10  4 y  11
 5x  9  1
 5x  10
x  2
y  2x  3
y  2 2   3
y  4  3  1
 2,1
10  11
False Stmt.
No Solution
2
x2
3
10 x  15 y  30
y
2

10 x  15 x  2   30
3

10x 10x  30  30
 30  30
True Stmt.
Infinite Solutions
But not done!
Answer should be a point
( x, mx + b )
2
x2
3
 2

 x, x  2 
 3

y
Sect 8.1 Systems of Linear Equations
Solve for x & y.
3x  4 y  1
2 x  3 y  12
Elimination Method.
1. Choose variable to cancel out.
Look for opposite signs.
2. Add the equations together to cancel.
3. Solve for the remaining variable.
4 Substitute this answer into either
equation in the step 1 equations.
Step 4
Step 1
The y-terms are opposite signs.
Multiply the first equation by 3
and the second equation by 4.
3x  4 y  13 
 9 x  12 y  3
Step 2
+
2 x  3 y  124 
 8 x  12 y  48
23  3 y  12
6  3 y  12
3y  6
y2
17 x  0 y  51
17 x 51 Step 3

17 17
x3
(3, 2) is the intersection
point and solution.
Solve the system for x and y.
The y-terms are
 2 5 x  4 y  22 
 2 5 x  7 y  29 
and easiest to cancel. To
multiples of 4. Multiply
determine which factor will
 3x  8 y  18 the 1st equation by -2 to
5 2 x  3 y  17  be negative check the y-terms
+
make opposites.
 10 x  8 y  44
 10 x  14 y  58
Add equations together.
+
10 x  15 y  85
13x
 26
y  27
x2
The x-terms are the smallest
52  4 y  22
10  4 y  22
4 y  12
y 3
2,3
2x  327  17
2x  81  17
2 x  64
x  32
 32,27
Solve the system for x and y.
Remove fractions by
2
3
35 x  y  5  multiply by the LCD and
decimals by multiply by 10’s
7
5
10 0.8 x  0.3 y  4.1
10 x  21 y  175
 8 x  3 y  41  7
+
10 x  21 y  175
56 x  21 y  287
66 x
 462
x7
107  21y  175
70  21y  175
21 y  105
y 5
7,5
2 5 x  8 y  11  10x  16 y  22
+
5 +2 x  8 y  13  10x  40 y  65
7x
 24
56 y  43
24
43
x
y
7
56
 24 
2   8 y  13
 7 
48
 8 y  13
7
43 1
1

 8y 
7 8
8
43
y
56
 24 43 
 , 
 7 56 
Solve the system for x and y.
2 3x  2 y  4
 6x  4 y  6
6x  4 y  8
0  14
False Stmt.
No Solution
8 x  2 y  4   2
 4x  y  2
4 x  y  2
00
True Stmt.
Infinite Solutions
Solve for y.
4 x  y  2
4x  2  y
Answer should be a point
( x, mx + b )
x,4x  2
Total-Relationship Systems.
In 2008, there were 746 species of plants that were considered threatened or
endangered. The number considered threatened was 4 less than a fourth of the
number considered endangered. How many plants are considered threatened and
endangered in 2008?
Find the two unknown’s from the question and determine their TOTAL.
Always read the question sentence first.
1
T + E = 746
4 E  4  E  746
T = How many threatened?  146
1.25E  4  746
E = How many endangered?  600
Now read through the details to
1.25E  750
T = ¼( E ) – 4
find the RELATIONSHIP between
1.25 1.25
the two variables.
Substitution Method
E  600


T  E  746
T  600  746
T  146
Total-Relationship System.
The sum is 90 o
Two angles are complementary. One angle is 12o less than twice the other.
Find the measure of the two angles.
Find the two unknown from the question and determine their TOTAL.
Always read the question sentence first.
2B 12  B  90
A
+
B
=
90
A = First angle  56
3B 12  90
B = Second angle  34
3B  102
A = 2( B ) – 12
Now read through the details to
3
3
find the RELATIONSHIP between
B  34
Substitution Method
the two variables.


A  B  90
A  34  90
A  56
*** If the two angles are supplementary, then the sum is 180 o
Total-Rate Systems.
A jewelry designer purchased 80 beads for a total of $39. Some of the beads
were silver beads that cost 40 cents each and the rest were gold beads that cost
65 cents each. How many of each type did the designer buy?
Find the two unknown from the question and determine their TOTAL.
Always read the question sentence first.
 0.40G  0.40S  32
 0.40 G + S = 80
G = How many Gold beads?  28
0.65G  0.40 S  39
S = How many Silver beads?  52
0.25G
7
Now read through the details to
0.65G + 0.40S = 39.00
0.25
0.25
find the RATE on each variable.
Multiply the rates to the variables Elimination Method
G  28
Cancel smallest variable term.
and set equal to total cost.


G  S  80
28  S  80
S  52
Total-Rate Systems.
Jane’s student loans total $9,600. She has a PLUS loan at 8.5% and a Stafford
loan at 6.8% simple interest. In one year, she was charged $729.30 in simple
interest. How much was each loan?
Find the two unknown from the question and determine their TOTAL.
Always read the question sentence first.
 0.068 P + S = 9600
P = How much was the PLUS loan?  $4500
S = How much was the Stafford loan?  $5100


 0.068P  0.068S  652.80
0.085P  0.068S  729.30
Now read through the details to
0.085P + 0.068S = 729.30
find the RATE on each variable.
Multiply the rates to the variables
Elimination Method
and set equal to the total cost.
Cancel smallest variable term.
Percent must be changed to a
decimal!
0.017P  76.50
0.017
0.017
P  4500
P  S  9600
4500  S  9600
S  5100
Total-Rate Systems.
A child ticket costs $3 and an adult ticket costs $5 at an afternoon movie. 300
tickets were sold for $1,150. How many of each type of ticket were purchased?
Find the two unknown from the question and determine their TOTAL.
Always read the question sentence first.
 3C  3 A  900
 3 C + A = 300
C = How many Child tickets?
 175
A = How many Adult tickets?  125


3C + 5A = 1150
Now read through the details to
find the RATE of each variable.
Multiply the rates to the variable
Elimination Method
and set equal to total cost.
Cancel smallest variable term.
3C  5 A  1150
2 A  250
2
2
A  125
A  C  300
125  C  300
C  175
Total-Mixture Systems.
Cashews cost $4 per pound and Walnuts cost $10 per pound. How much of each
type should be used to make a 50 pound mixture that sells for $5.80 per pound?
Find the two unknown from the question and determine their TOTAL.
Always read the question sentence first.
C = How many pounds of Cashews?  35
W = How many pounds of Walnuts?  15


 4 C + W = 50
 4C  4W  200
4C  10W  290
Now read through the details to
4C + 10W = 5.80(50)
find the RATE of each variable and
TOTAL. Multiply the rates to the
Elimination Method
variable and TOTAL.
Cancel smallest variable term.
6W  90
6
6
W  15
C  W  50
C  15  50
C  35
Total-Mixture Systems.
Home Depot carries two brands of liquid fertilizers containing nitrogen and
water. Gentle Grow is 3% nitrogen while Super Grow is 8% nitrogen. Home
Depot needs to combine the two types of solution to fill a customer’s order that
requested 90L of fertilizer that is 6% nitrogen. How much of each brand should
be used to fill the order?
Find the two unknown from the question and determine their TOTAL.
Always read the question sentence first.
G = How many liters of Gentle Grow?  0.03 G + S = 90
 36
S = How many liters of Super Grow?
 54


 0.03G  0.03S  2.7
0.03G  0.08S  5.4
0.03G + 0.08S = 0.06(90)
Now read through the details to
find the RATE of each variable and
Elimination Method
TOTAL. Multiply the rates to the
Cancel smallest variable term.
variable and TOTAL.
0.05S  2.7
0.05
0.05
S  54
G  S  90
G  54  90
G  36
Distance Systems.
A jet flies 4 hours west with a 60 mph tailwind. Returning against the same
wind, the jet takes 5 hours. What is the speed of the jet with no wind?
D
With
wind
D
Against
wind
D
=
r
*
r  60
r  60
t
4
 D  4r  240
Substitution Method
5
 D  5r  300
4r  240  5r  300
 4r
 4r
240  r  300
 300   300
540  r
540 mph in no wind.
Distance Systems.
A freight train leaves Chicago heading to Denver at a speed of 40 mph. Two hours
later an Amtrak train leaves Chicago bound for Denver at a speed of 60 mph.
How far will the trains travel until the Amtrak passes the freight train?
D
=
r
*
t
Freight
train
D
40
t
Amtrak
D
60
t  2
 D  40t
Substitution Method
 D  60t 120
D  40t
40t  60t 120
 60t  60t
D  406 
 20t  120
D  240
 20  20
t 6
They will travel 240 miles.