the numerator must always be 1 degree less than the denominator. a, b, c,..., and z are constants
TRANSCRIPT
The numerator must always be 1 degree less than the denominator. A, B, C,..., and Z are constants.
The numerator must always be 1 degree less than the denominator. That is why there are 2 terms on the top.
Write the partial fraction decomposition of xxx
x
23 2
2
Factor the denominator.
2
2
2 1
1 1 1
x x x
x x x x x
Write a fraction for each factor. The factors are all LINEAR!
23 2
2
2 1 1
x A B C
x x x x x x
1
Every fraction needs the LCDMultiply top and bottom by what is missing.
21x x
2
2 2 23 2
1 12
2 1 1 1
A x Bx xx Cx
x x x x x x x x x
Multiply out the tops and remove the bottoms, they are all the same.
2 22 2 1x Ax Ax A Bx Bx Cx Collect all like terms.
0x2 +
x2 terms: 0 = A + B
x terms: 1 = -2A – B + C
Constants: 2 = A
Back substitute.
0 = 2 + B-2 = B
1 = -2(2) – (-2) + C1 = -2 + C3 = C
23 2
2 2 2 3
2 1 1
x
x x x x x x
Substitute the values in for A, B, and C.
2 X
Modified Zero Product Rule, another way to find A, B, and C.
22 1 1x A x Bx x Cx
Let x = 0 20 2 0 1
2
A
A
Let x = 1 1 2 1
3
C
C
Let x = 2
22 2 2 1 2 2 1 2
4 2 2
4 2 2 2 3
A B C
A B C
B
4 2 8
4 2
2
B
B
B
Write the partial fraction decomposition of 2 2
4
4
x
x x
The denominator is factored.
Write a fraction for each factor.
22 2 2
4
4 4
x A B Cx D
x xx x x
Every fraction needs the LCDMultiply top and bottom by what is missing.
2 2 4x x
Multiply out the tops and remove the bottoms, they are all the same.
3 2 3 24 4 4x Ax Ax Bx B Cx Dx
Collect all like terms.
x2 terms: 0 = B + D x terms: 1 = 4A; A = ¼
Constants: 4 = 4B; B = 1
Back substitute.
0 = ¼ + C- ¼ = C
0 = 1 + D-1 = D
Modified Zero Product Rule
Let x = 0
Linear Quadratic
x3 terms: 0 = A + C
44
4
4
4
4
422
2
22
2
22
2
22
xx
xDCx
xx
xB
xx
xAx
xx
x
222 444 xDCxxBxAxx
B
B
B
1
44
040040 2
4
14114
1
4
42222
x
x
xxxx
x
4
14114
1
4
42222
x
x
xxxx
x
Doesn’t work well with Quad. Factors!
( )(-1)
23 yx+
232 2 xx
0232 2 xx
0212 xx
2;2
1
xx
23 yx
2x2
1x
y
y
y
8
26
223
y
y
y
2
1
22
3
22
13
23 yx
2
1,
2
1
8,2
Solve by Elimination.Back Substitute to find y.
Calculator Check. Set = to y.
7
132
22
yx
yx
Solve the following systems of equations:
( )(-1)
72 yx+ 1322 yx
Solve by Elimination.
62 yy
062 yy
Solve y.
2;3
023
yy
yy
Back Substitute to find x.
2y3y
72 yx 72 yx
2
4
73
73
2
2
2
x
x
x
x
3
9
722
2
x
x
x
3,2&3,2 2,3&2,3 Calculator Check. Set = to y.Graph, 2nd Trace, Intersection
Zoom 6 Standard Window
Zoom 5 Square Window
1st Curve? Y12nd Curve? Y31st Curve? Y12nd Curve? Y3
1st Curve? Y22nd Curve? Y31st Curve? Y22nd Curve? Y3
182
422
yx
xy
Solve the following systems of equations:Solve by Substitution.
xy
4
184
22
2
x
x
24
22
18162
1816
2
xx
xx
018
089
016182
22
24
24
xx
xx
xx
22
8
082
2
x
x
x
1
1
012
2
x
x
x
xy
4
xy
4
2,22
2,22
4
24,22
2
2
22
4,22
22
4,22
4,11
4,1
4,11
4,1
Multiply all terms by x2
Set = 0 and write in descending order.
Divide by2.Factor like a quadratic.
Calculator check set up for Y =
Solve the following systems of equations:
24log
32log
y
y
x
x
Identify restrictions on answers for x and y.
x is a base… x > 0 and x can’t = 1.
y is in the log … y > 0 .
Convert both equations to exponential.
yx
yx
4
22
3
Solve by Substitution.
yx
2
3
24
32 xx
120
20
2
2
23
32
xx
xx
xx
2
1;0 xx
16
1
8
1
2
1
2
1
2
1
2
1
2
33
3
x
xy
16
1,
2
1
19Section 8.7Systems of Inequalities.There will either be _________ solutions or ______________ solutions.
Solve the following systems of equations:
02
42
yx
yx
NO INFINITE
42 yx 02 yxFind x and y intercepts.
( _____ , 0 )
( 0 , _____)
4
40
x
x
2
420
y
y
4
2
Plot the intercepts and draw the line.
Test the origin to see if the inequality is true. If it is true, we shade in that direction. If not true, we shade away from the origin.
TRUESolve for y
xy 2Graph like y = mx + b When solved for y, the > symbol tells us to shade above the line. If we have a < symbol, then we shade below the line. Remember…if there is no equal to line, then the line is a dashed line.
2 ways to graph.
The solution is the OVERLAP
19
Solve the following systems of equations:
5
252
22
xy
yx
Identify the graphs.
Circle
Parabola
2522 yx r = 5, Center @ ( 0, 0 )
The < symbol means we want all the points inside the circle. The > symbol means we want to shade all the points outside the circle.
52 xy Vertex at ( 0, -5 ) with our 1, 3, 5, ... pattern.
y < means we shade below the curve.
The solution is the OVERLAP
19Solve the following systems of equations:
62
62
0
0
yx
yx
y
x
62 yx 62 yx
Find the x and y intercepts.
)0,3(
3
62
x
x
)6,0(
6y
)0,6(
6x
)3,0(
3
62
y
y
Ax + By = C; Standard Form Note.
The slope is m = -A / B
Write the system of inequalities.
Take care of the easy lines first, horizontal and vertical lines.
3x
2y
4y
42 xy
The slope is m = -2/1
Y-intercept (0, 4), b = 4
y = -2x + 4
0
0
2 4
1
x
y
x y
x y
Subject to the following constraints.
Find the Maximum value ofz = 3x + 2y
11
11
xy
xy1 2 3 54
1
2
3
5
4
( 2, 1 )
( 0, 2 )
( 1, 0 )
z = 3x + 2y
( 0, 2 ): z = 3(0) + 2(2) = 4
( 2, 1 ): z = 3(2) + 2(1) = 8
( 1, 0 ): z = 3(1) + 2(0) = 3
( 2, 1 ) gives the maximum value of 8.
0
0
2 3 6
3 15
4
2 5 27
x
y
x y
x y
x y
x y
Find the maximum and minimum value of z = 5x + 7y
Subject to the following constraints
Quad 1 only.
Max: Points farthest from the origin.
Find intercepts.
Find point & slope.
1
3
1
3;0,5
m
1
1;4,0 m
Intercepts are fractions…To find a point, use multiples of 2 and 5 to get a sum that is 27. 2 + 25 = 27. Therefore x = 1 and y = 5 for this to work. Find the slope and graph the line.
5
2;5,1
m
62 84 10 12
6
2
8
4
10
12
Min: Points closest to the origin.
( 1, 5 )
( 6, 3 )
( 1, 5 ): z = 5(1) + 7(5) = 40( 6, 3 ): z = 5(6) + 7(3) = 51( 6, 3 ) is the point for max.
( 0, 2 ): z = 5(0) + 7(2) = 14
( 0, 2 )( 3, 0 )
( 3, 0 ): z = 5(3) + 7(0) = 15( 0, 2 ) is the point for min.