Rational-Functions-GENERAL MATHEMATICS.pptx

MANUEL S. ENVERGA UNIVERSITY FOUNDATION
An Autonomous University
BASIC EDUCATION DEPARTMENT
Session 2
RATIONAL FUNCTIONS
MSEUF, Lucena City
April 18, 2017
Facilitator: Mr. WILLIAM M. VERZO

At the end of the session, the teacher-participants are expected to:
1. represents real-life situations using rational functions.
2. distinguishes rational function, rational equation, and rational
inequality.
3. solves rational equations and inequalities.
4. represents a rational function through its: (a) table of values, (b) graph,
and (c) equation.
5. finds the domain and range of a rational function.
6. determines the: (a) intercepts; (b) zeroes; and asymptotes of rational
functions.
7. graphs rational functions.
8. solves problems involving rational functions, equations, and
inequalities.

PRIMING ACTIVITY
1. 3x2
-8x+4
2. 11x2
-99
3. 16x3
+128
Factor the following completely:
4. x3
+2x2
-4x-8
5. 2x2
-x-15
6. 10x3
-80
(3x-2)(x-2)
11(x+3)(x-3)
16(x+2)(x2
-2x+4)
(x-2)(x+2)2
(2x+5)(x-3)
10(x-2)(x2
+2x+4)

PRIMING ACTIVITY
• Solve the following rational equation.
x
x
x
x
x
2
4
1
4
12
2





: 1
: 4
Solution x
Extr x



PRIMING ACTIVITY
0
8
2
12 2




 x
x
x
0
4
3
2



 x
x
0
4
3
2


 x
x
0
)
1
)(
4
( 

 x
x
1
4 

 x
or
x
x = -4 Extraneous
x = 1

ACTIVITY 1 (Group Work) (15 min)
1. Your task is to complete the table to show that the time it takes to
reach the top of wall depends on the climber’s speed.
2. Compare your results and describe their properties.
3. The team leaders of the groups report their conclusions to the whole
class.
Climbing the Wall

ANALYSIS
1. What did you find difficult about this task?
2. What task did you find most difficult to do? Why?
3. What information can you get from the equation of a rational graph?
4. What have you learned about the key features of the rational function?
5. What are some common errors which students may commit? How can
we prevent such error? Feel free to offer a suggestion.

ABSTRACTION
LEARNING AREA STANDARD:
At the end of the course, the students must know how to solve problems
involving rational, exponential and logarithmic functions; to solve business-
related problems; and to apply logic to real-life situations.
CONTENT STANDARD:
The learner demonstrates understanding of key concepts of rational
functions.

ABSTRACTION
PERFORMANCE STANDARD:
The learner is able to accurately formulate and solve real-life problems
involving rational functions.

ABSTRACTION
LEARNING COMPETENCIES:
The learners ...
a. represents real-life situations using rational functions.
b. distinguishes rational function, rational equation, and rational
inequality.
c. solves rational equations and inequalities.
d. represents a rational function through its: (a) table of values, (b) graph,
and (c) equation.
e. finds the domain and range of a rational function.

ABSTRACTION
LEARNING COMPETENCIES:
The learners ...
f. determines the: (a) intercepts; (b) zeroes; and asymptotes of rational
functions.
g. graphs rational functions.
h. solves problems involving rational functions, equations, and
inequalities.

Many real-world problems can be modeled by rational functions.
REPRESENTING REAL LIFE SITUATIONS
USING RATIONAL FUNCTIONS

REPRESENTING REAL LIFE SITUATIONS
USING RATIONAL FUNCTIONS

RATIONAL FUNCTIONS
Rational Expression
 It is the quotient of two polynomials.
 A rational function is any ratio of two polynomials, where
denominator cannot be ZERO!
Examples:
2
5
x
y
x



2
3 2
3 2 5
4 5 7
x x
y
x x x
 

  
Not Rational:
4
2
x
y
x

 2
5
x
y
x



RATIONAL FUNCTIONS
Asymptotes
 Asymptotes are the boundary lines that a rational function
approaches, but never crosses.
 We draw these as Dashed Lines on our graphs.
 There are three types of asymptotes:
Vertical
Horizontal (Graph can cross these)
Slant

RATIONAL FUNCTIONS
Vertical Asymptotes
 Vertical Asymptotes exist where the denominator would be zero.
 They are graphed as Vertical Dashed Lines
 There can be more than one!
 To find them, set the denominator equal to zero and solve for “x”

RATIONAL FUNCTIONS
Vertical Asymptotes
1
)
(


x
x
x
f
•Set the denominator equal to zero
•x – 1 = 0, so x = 1
•This graph has a vertical asymptote
at x = 1
• Find the vertical asymptotes for the
following function:

1 2 6
3 4 5 7 8 9 1
0
4
3
2
7
5
6
8
9
x-
axis
y-
axis
0
1
-2
-6 -3
-4
-5
-7
-8
-9
1
0
-4
-3
-2
-1
-7
-5
-6
-8
-9
0
-1
Vertical
Asymptote at
X = 1

Other Examples:
• Find the vertical asymptotes for the following
functions:
3
3
)
(


x
x
g
)
5
)(
2
(
1
)
(




x
x
x
x
g
3
: 
x
VA
5
:
2
:



x
VA
x
VA

Horizontal Asymptotes
• Horizontal Asymptotes are also Dashed Lines drawn
horizontally to represent another boundary.
• To find the horizontal asymptote you compare the degree
of the numerator with the degree of the denominator

Horizontal Asymptote (HA)
Given Rational Function:
Compare DEGREE of Numerator to Denominator
If N < D , then y = 0 is the HA
If N > D, then the graph has NO HA
If N = D, then the HA is
Numerator
( )
Denominator
f x 
N
D
LC
y
LC


Example #1
• Find the horizontal asymptote for the
following function:
1
)
(


x
x
x
f
•Since the degree of numerator is equal
to degree of denominator (m = n)
•Then HA: y = 1/1 = 1
•This graph has a horizontal asymptote
at y = 1

1 2 6
3 4 5 7 8 9 1
0
4
3
2
7
5
6
8
9
x-
axis
y-
axis
0
1
-2
-6 -3
-4
-5
-7
-8
-9
1
0
-4
-3
-2
-1
-7
-5
-6
-8
-9
0
-1
Horizontal
Asymptote at
y = 1

Other Examples:
• Find the horizontal asymptote for the following
functions:
3
3
)
(


x
x
g
1
3
1
3
)
( 2
2




x
x
x
x
g
5
1
)
(
3



x
x
x
g
0
: 
y
HA
3
: 
y
HA
None
HA :

Slant Asymptotes (SA)
• Slant asymptotes exist when the degree of the numerator is
one larger than the denominator.
• Cannot have both a HA and SA
• To find the SA, divide the Numerator by the Denominator.
• The results is a line y = mx + b that is the SA.

Example of SA
27
2
2 4 8
( )
2
x x
f x
x
 


2 4 8
 
-2
2
4

8

16
8
2 8
y x
 
Remainder does not matter

28
Holes
• A hole exists when the same factor
exists in both the numerator and
denominator of the rational
expression and the factor is
eliminated when you reduce!

Example of Hole Discontinuity
29
( 4)( 1)
( )
( 2)( 4)
x x
f x
x x
 

 
Cancel LIKE factors
( 1)
( )
( 2)
x
f x
x



 
____, ____
4

( 4 1) 5 5
( 4 2) 6 6
  
 
  
5
6

3.5 - 30
Domain: (–, 0)  (0, ) Range: (–, 0)  (0, )
Find domain and graph.
x y
–2 –½
–1 –1
–½ –2
0 undefined
½ 2
1 1
2 ½
x
x
f
1
)
( 
 It is discontinuous at x = 0.

3.5 - 31
Domain: (–, 0)  (0, ) Range: (–, 0)  (0, )
x y
–2 –½
–1 –1
–½ –2
0 undefined
½ 2
1 1
2 ½
 decreases on the intervals
(–,0) and (0, ).
1
( )
x
x

f
x
x
f
1
)
( 

3.5 - 32
Domain: (–, 0)  (0, ) Range: (–, 0)  (0, )
x y
–2 –½
–1 –1
–½ –2
0 undefined
½ 2
1 1
2 ½
x
x
f
1
)
( 
 The y-axis is a vertical
asymptote, and the x-axis is
a horizontal asymptote.

3.5 - 33
Domain: (–, 0)  (0, ) Range: (–, 0)  (0, )
x y
–2 –½
–1 –1
–½ –2
0 undefined
½ 2
1 1
2 ½
x
x
f
1
)
( 
 It is an odd function and its
graph is symmetric with respect
to the origin.

3.5 - 34
Domain: (–, 0)  (0, ) Range: (0, )
x y
 3
 2 ¼
 1 1
 ½ 4
 ¼ 16
0 undefined
 increases on the interval (–,0) and
decreases on the interval (0, ).
2
1
( )
x
x

f
1
9
2
1
( )
x
x

f

3.5 - 35
Domain: (–, 0)  (0, ) Range: (0, )
x y
 3
 2 ¼
 1 1
 ½ 4
 ¼ 16
0 undefined
It is discontinuous at x = 0.
1
9
2
1
( )
x
x

f

3.5 - 36
Domain: (–, 0)  (0, ) Range: (0, )
x y
 3
 2 ¼
 1 1
 ½ 4
 ¼ 16
0 undefined
 The y-axis is a vertical asymptote, and the x-axis is a
horizontal asymptote.
1
9
2
1
( )
x
x

f

3.5 - 37
Domain: (–, 0)  (0, ) Range: (0, )
x y
 3
 2 ¼
 1 1
 ½ 4
 ¼ 16
0 undefined
 It is an even function, and Its graph is symmetric with
respect to the y-axis.
1
9
2
1
( )
x
x

f

RATIONAL EQUATIONS

RATIONAL INEQUALITIES

RATIONAL FUNCTIONS
Lesson 4: Representations of Rational Functions
Lesson 4:

GRAPHING RATIONAL FUNCTIONS
General Steps to Graph a Rational Function
1) Factor the numerator and the denominator
2) State the domain and the location of any holes in the graph
3) Simplify the function to lowest terms
4) Find the y-intercept (x = 0) and the x-intercept(s) (y = 0)
5) Identify any existing asymptotes (vertical, horizontal, or
oblique

GRAPHING RATIONAL FUNCTIONS
6) Identify any points intersecting a horizontal or oblique asymptote.
7) Use test points between the zeros and vertical asymptotes to locate
the graph above or below the x-axis
8) Analyze the behavior of the graph on each side of an asymptote
9) Sketch the graph

The Graph of a Rational Function
y-intercept (x = 0) x-intercept(s) (y = 0)
𝑓 (0)=
(0+4)(0− 3)
(0+2)(0−2)
𝑓 (0)=
− 12
− 4
=3
(0 ,3)
Use numerator factors
𝑥+4=0 𝑥−3=0
𝑥=−4 𝑥=3
(−4 ,0) (3 ,0)

oblique
Horiz. Or Oblique Asymptotes Vertical Asymptotes
𝑦 =
1
1
𝐻𝐴: 𝑦=1
Use denominator factors
𝑥+2=0 𝑥−2=0
𝑥=−2 𝑥=2
𝑉𝐴:𝑥=−2𝑎𝑛𝑑 𝑥=2
𝑓 (𝑥 )=
𝑥2
+ 𝑥 −12
𝑥
2
− 4
𝑓 ( 𝑥)=
(𝑥 +4 )( 𝑥 −3)
(𝑥+ 2)(𝑥 −2)
Examine the largest exponents
Same Horiz. - use coefficients

6) Identify any points intersecting a horizontal or oblique
asymptote.
𝑦 =1 𝑎𝑛𝑑 𝑓 ( 𝑥) =
𝑥2
+ 𝑥 − 12
𝑥
2
− 4
1=
𝑥 2
+ 𝑥 − 12
𝑥
2
− 4
𝑥 2
− 4 = 𝑥 2
+ 𝑥 − 12
− 4 = 𝑥 − 12
8 = 𝑥
( 8,1 )

𝑓 (𝑥)=
(𝑥+4)(𝑥 −3)
(𝑥+2)(𝑥− 2)
7) Use test points between the zeros and vertical asymptotes
to locate the graph above or below the x-axis
-4 -2 2 3
𝑓 (−5)=
(−5+4)(−5− 3)
(−5+2)(−5−2)
𝑓 (−5)=
(−)(−)
(−)(−)
=+¿
𝑓 (−5)=𝑎𝑏𝑜𝑣𝑒
𝑓 (−3 )=¿¿
𝑓 (−3 )=𝑏𝑒𝑙𝑜𝑤
𝑓 (0)=¿ ¿
𝑓 (0)=𝑎𝑏𝑜𝑣𝑒
𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤 𝑎𝑏𝑜𝑣𝑒

𝑓 (𝑥)=
(𝑥+4)(𝑥 −3)
(𝑥+2)(𝑥− 2)
-4 -2 2 3
𝑓 (2.5)=¿¿
𝑓 (2.5)=𝑏𝑒𝑙𝑜𝑤
𝑓 (4 )=¿ ¿
𝑓 (4 )=𝑎𝑏𝑜𝑣𝑒
𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤 𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤 𝑎𝑏𝑜𝑣𝑒

𝑓 (𝑥)=
(𝑥+4)(𝑥 −3)
(𝑥+2)(𝑥− 2)
𝑥 → −2−
𝑎𝑏𝑜𝑣𝑒
-4 -2 2 3
𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤 𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤
8) Analyze the behavior of the graph on each side of an
asymptote
𝑓 (𝑥)→¿¿ 𝑓 (𝑥)→−∞
𝑥→−2
+¿ ¿
𝑓 (𝑥)→¿¿ 𝑓 (𝑥)→∞

𝑓 (𝑥)=
(𝑥+4)(𝑥 −3)
(𝑥+2)(𝑥− 2)
𝑥 → 2−
asymptote
𝑓 (𝑥)→¿¿ 𝑓 (𝑥)→∞
𝑥→2
+¿ ¿
𝑓 (𝑥)→¿¿ 𝑓 (𝑥)→−∞
-4 -2 2 3
𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤 𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤

9) Sketch the graph

Example
𝑓 ( 𝑥 ) =
( 𝑥 − 2 )
( 𝑥 + 3 )
𝑓 ( 𝑥 ) =
𝑥 2
+3 𝑥 − 10
𝑥
2
+8 𝑥 +15
𝑓 ( 𝑥 ) =
( 𝑥 +5 ) (𝑥 − 2 )
( 𝑥 + 5 )( 𝑥 + 3 )
Domain:
Hole in the graph at

𝑓 (0)=
(0 − 2)
(0+3)
𝑓 (0)=−
2
3
(0 ,−
2
3
)
𝑥−2=0
𝑥=2
(2,0)

oblique
Horiz. Or Oblique Asymptotes Vertical Asymptotes
𝑦 =
1
1
𝐻𝐴: 𝑦=1
𝑥+3=0
𝑥=−3
𝑉𝐴: 𝑥=−3
𝑓 (𝑥 )=
𝑥2
+3 𝑥 −10
𝑥
2
+8 𝑥 +15
𝑓 (𝑥 )=
( 𝑥 −2)
( 𝑥+ 3)
Same Horiz. - use coefficients

asymptote.
𝑦 =1 𝑎𝑛𝑑 𝑓 ( 𝑥 ) =
𝑥 − 2
𝑥 +3
1=
𝑥 − 2
𝑥 + 3
𝑥 + 3 =𝑥 − 2
3 = − 2
𝑙𝑜𝑠𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
𝑛𝑜𝑝𝑜𝑖𝑛𝑡𝑠𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑜𝑛 h
𝑡 𝑒𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒

𝑓 (𝑥 )=
( 𝑥 −2)
( 𝑥+ 3)
𝑓 (− 4)=
(− 4 − 2)
(− 4 +3)
𝑓 (− 4)=
(−)
(−)
=+¿
𝑓 (− 4)=𝑎𝑏𝑜𝑣𝑒
𝑓 ( 0) =
(−)
¿ ¿
𝑓 (0)=𝑏𝑒𝑙𝑜𝑤
𝑓 (3)=¿ ¿
𝑎𝑏𝑜𝑣𝑒 𝑏𝑒𝑙𝑜𝑤 𝑎𝑏𝑜𝑣𝑒
-3 2

𝑓 (𝑥 )=
( 𝑥 −2)
( 𝑥+ 3)
𝑥 → −3−
asymptote
𝑓 (𝑥)→
(−)
(0
−
)
𝑓 (𝑥)→∞
𝑥→−3
+¿¿
𝑓 (𝑥)→
(−)
¿¿
𝑓 (𝑥)→−∞
-3 2

Example
𝑓 ( 𝑥 ) =
( 𝑥 + 2)( 𝑥 +1 )
( 𝑥 − 1 )
𝑓 ( 𝑥 ) =
𝑥 2
+3 𝑥 + 2
𝑥 − 1
𝑓 ( 𝑥 ) =
( 𝑥 + 2)( 𝑥 +1 )
𝑥 − 1
Domain:
No holes

𝑓 (0)=
(0 +2)(0+1)
(0 − 1)
𝑓 (0)=
2
− 1
=−2
(0 ,−2)
𝑥+2=0
𝑥=−2
(−2,0)
𝑥+1=0
𝑥=−1
(−1,0)

oblique
Horiz. or Oblique Asymptotes Vertical Asymptotes
𝑥−1
O 𝐴: 𝑦=𝑥+4
𝑥−1=0
𝑥=1
𝑉𝐴: 𝑥=1
𝑓 (𝑥 )=
𝑥2
+3 𝑥+ 2
𝑥 − 1
𝑓 (𝑥 )=
(𝑥 +2)(𝑥 +1)
(𝑥 −1)
Oblique: Use long division
2
3
2

 x
x
𝑥
𝑥2
− 𝑥
−+¿
4 𝑥
+2
+4
4 𝑥− 4
−+¿
0

asymptote.
𝑦 =𝑥+ 4 𝑎𝑛𝑑 𝑓 (𝑥 )=
( 𝑥+2)( 𝑥 +1)
𝑥 −1
𝑥 + 4 =
(𝑥 +2 )( 𝑥 +1)
𝑥 − 1
( 𝑥+ 4 )( 𝑥 −1)=( 𝑥+ 2)( 𝑥 +1)
𝑥 2
+ 3 𝑥 − 4 = 𝑥2
+3 𝑥 +2
𝑙𝑜𝑠𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
𝑛𝑜𝑝𝑜𝑖𝑛𝑡𝑠𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑜𝑛 h
𝑡 𝑒𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒

𝑓 (𝑥)=
(𝑥+2)(𝑥+1)
(𝑥 −1)
𝑓 (− 4)=
(−)(−)
(−)
=−
𝑓 (−1.5)=¿¿
𝑓 (− 4)=𝑏𝑒𝑙𝑜𝑤 𝑓 (0)=¿¿
𝑓 (0)=𝑏𝑒𝑙𝑜𝑤
𝑓 (3)=¿¿
𝑏𝑒𝑙𝑜𝑤 𝑎𝑏𝑜𝑣𝑒
-2 1
-1
𝑓 (−1.5)=𝑎𝑏𝑜𝑣𝑒
𝑏𝑒𝑙𝑜𝑤

𝑓 (𝑥 )=
(𝑥+2)(𝑥 +1)
(𝑥 −1)
𝑥 → 1−
asymptote
𝑓 (𝑥)→¿¿ 𝑓 (𝑥)→−∞
𝑥 → 1
+¿¿
𝑓 (𝑥)→¿¿ 𝑓 (𝑥 ) → ∞
1

Review: STEPS for GRAPHING
HOLES
___________________________________________
___________________________________________
EX _________________________________________
EX _________________________________________
Discontinuous part of the graph where the line jumps over.
Represented by a little open circle.
)
5
x
)(
3
x
(
)
3
x
(
y




)
2
x
(
x
)
2
x
(
x
y 2



Hole @ x = 3
Hole @ x = 2
No hole at x = 0

VERTICAL ASYMPTOTES
___________________________________________
___________________________________________
EX _________________________________________
EX _________________________________________
Discontinuous part of the graph where the line cannot cross over.
Represented by a dotted line called an asymptote.
)
2
x
(
)
5
x
(
y



)
5
x
)(
2
x
(
x
x
y



VA @ x = 2
Hole @ x =0
VA @ x = 2, -5

HORIZONTAL ASYMPTOTES
n = degree of numerator
d = degree of denominator
_______________________________________________
_______________________________________________
_______________________________________________
Case 1 n > d
)
2
x
(
7
x
5
y
2


 No HA
Case 2 n < d
1
x
3
x
y 3


 HA @ y = 0
Case 1 n = d
)
2
x
)(
2
x
(
5
1
x
4
y
2



 HA is the ratio of coefficients
HA @ y = 4 / 5

Finding holes and asymptotes
VA: x=-1, -5
HA: y=0 (power of the denominator
is greater than the numerator)
Holes: none
VA: none (graph is the same as
y=x-1 once the (x-2)s cancel
HA: none (degree of the numerator
is greater than the denominator)
Hole: x=2

Let’s try some
VA: x=3
HA: none (power of the numerator is
greater than the denominator)
Holes: x=2
VA: x=-5,0 ( cancel the (x-3)s
HA: y=0 (degree of the denominator
Hole: x=3
Find the vertical, horizontal asymptotes and any holes

GRAPHING
y = x / (x – 3)
1) HOLES?
no holes since nothing cancels
2) VERTICAL ASYMPTOTES?
Yes ! VA @ x =3
4) T-CHART
X Y = x/(x – 3)
4 Y = 4
2 Y = -2
3) HORIZONTAL ASYMPTOTES?
Yes ! HA @ y =1
0
5
Y = 0
Y = 5/2

GRAPHING
1) HOLES?
4) The graph -
What cancels?
Graph the function
y=x with a hole
at x=-1
hole @ x = -1
None!
None!

GRAPHING
1) HOLES?
4) T-CHART
X
6 Y = 1/2
-3 Y = -5/8
1
2
Y = 1/12
Y = 0
)
5
x
)(
2
x
(
x
)
2
x
(
x
y




)
5
x
)(
2
x
(
)
2
x
(
y




)
5
x
)(
2
x
(
)
2
x
(
y




3 Y = -1 / 10
WAIT –
What about
the
Horizontal
Asymptote
here?
hole @ x = 0
Yes ! VA @ x =-2 , 5
Yes ! HA @ y =0 (Power of the denominator

Remember,
Horizontal
Asymptotes only
describe the ends of
the function (left and
right). What happens
in the middle is ‘fair
game’.
T-CHART
X
-1 Y = 1/2
4 Y = -1/3
2 Y = 0
)
5
x
)(
2
x
(
)
2
x
(
y




To find out what the graph looks like between the
vertical asymptotes, go to a T Chart and plug in
values close to the asymptotes.
Left
Right
Middle

Let’s try one:
Sketch the Graph
1) HOLES?
4) T-CHART
X
0 Y = 0
-1 Y = 1/4
-2
2
Y = .22
Y=-2
3 Y = -3/4
none
Yes ! VA @ x = 1
Yes ! HA @ y =0 (Power of the denominator

Problems
Find the vertical asymptotes, horizontal asymptotes, slant
asymptotes and holes for each of the following functions.
 
2
2
2 15
7 10
x x
f x
x x
 

 
Vertical: x = -2
Horizontal : y = 1
Slant: none
Hole: at x = - 5
 
2
2 5 7
3
x x
g x
x
 


Vertical: x = 3
Horizontal : none
Slant: y = 2x +11
Hole: none

APPLICATION
On a manila paper, paste/draw some small pictures of objects such that
they are positioned at different coordinates.
Then, draw circles that contain these pictures.
Using the pictures and the circles drawn on the grid, formulate problems
involving the equation of the circle and then solve them.

CLOSURE
“The ability to simplify means to eliminate
the unnecessary so that the necessary may
speak.”
Hans Hofmann – early 20th century teacher and painter

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