EECS1022 Winter 2018 Additional Notes Tracing Point, PointCollector, and PointCollectorTester

Size: px
Start display at page:

Download "EECS1022 Winter 2018 Additional Notes Tracing Point, PointCollector, and PointCollectorTester"

Transcription

1 EECS1022 Winter 2018 Additional Notes Tracing, Collector, and CollectorTester Chen-Wei Wang Contents 1 Class 1 2 Class Collector 2 Class CollectorTester 7 1 Class 1 class { 2 double ; double ; 4 (double n, double n) { 5 = n; 6 = n; 7 } 8 } Figure 1: Template for Two-Dimensional s Figure 1 (page 1) shows the complete definition of class, which is a template for entities (i.e., ) on a two-dimensional plane. The common attributes (Lines 2 and ) of all are the -coordinate and -coordinate: 2 double ; double ; To create an instance of the above class, ou need to make a call to its (in this case, the onl) constructor (Lines 4 to 7) b giving two argument values whose tpes match the tpes of the constructor s two parameters. 4 (double n, double n) { 5 = n; 6 = n; 7 } For eample, the call (, 4) is valid, because the first argument value matches the tpe of the first parameter, and the second argument value should 4 matches the tpe of the second parameter. 1

2 2 Class Collector 1 class Collector { 2 [] ; int nop; /* number of */ 4 Collector() { 5 = new [100]; 6 } 7 void add(double, double ) { 8 [nop] = new (, ); 9 nop++; 10 } 11 [] getsinquadranti() { 12 [] ps = new [nop]; 1 int count = 0; /* number of in Quadrant I */ 14 for(int i = 0; i < nop; i ++) { 15 p = [i]; 16 if(p. > 0 && p. > 0) { 17 ps[count] = p; 18 count ++; 19 } 20 } 21 [] q1s = new [count]; 22 /* ps contains if count < nop */ 2 for(int i = 0; i < count; i ++) { 24 q1s[i] = ps[i] 25 } 26 return q1s; 27 } 28 } Figure 2: Template for Entities which Collect s Attributes of Collector : Figure 2 (page 2) shows the complete definition of class Collector, which is a template for entities (i.e., collectors of ), each of which collecting a list of on the two-dimensional plane. The common attributes (Lines 2 and ) of all point collectors are: Line 2: An arra of, which is of tpe []. Declaring the attribute [] ; means that at runtime, each point collector has an arra named, where each slot stores the address (which ma be ) of some object. Line : An integer counter named nop (standing for number of ), which serves two purposes: 1. Records how man have been collected (i.e., added to the arra) so far. Notice that the value of nop should start with 0, and get incremented b one as each new object is collected, until it reaches.length. 2. Indicates where in (i.e., an inde value between 0 and.length - 1). When the value of nop reaches its maimum.length, meaning that all slots of the arra 2

3 are occupied b addresses of objects, this maimum value also means that storing a net object is not possible (because the value.length is out of the inde bounds for arra ). Constructor of Collector : To create an instance of the above Collector class, ou need to make a call to its (onl) constructor (Lines 4 to 6) b giving zero argument values. For eample, consider the the following line Collector pc = new Collector(); where Step 1: The RHS (i.e., new Collector()) creates a new object of tpe PersonCollector. Step 2: The LHS (i.e., Collector pc) declares a variable pc which, at runtime, stores onl an address of some object. Step : The middle equal sign (=) assigns (i.e., stores) the address of the new object (created in Step 1) to the address placeholder (declared in Step 2). Here is a visual summar of the effect of eecuting the above three steps (where we use an addition dashed arrow to make it eplicit that the value of nop can be seen as pointing to a position in arra pc.): Object Structure After Initializing the Collector pc nop == 0 Collector where The path pc.nop (where pc is the contet object) brings us to the value of attribute nop of the specific Collector instance pc. The path pc.ponts (where pc is the contet object) brings us to the value of attribute. That is, pc. denotes the address of an arra object, whose each slot stores the address of some object. Wh is it the case that pc.nop has the value 0 and pc. has the value of an arra whose each slot stores the (i.e., unknown address) value? Inspect the definition of the constructor in class Collector (Lines 5 to 6 in Figure 2, page 2): 4 Collector() { 5 = new [100]; 6 }

4 When each instance of tpe Collector is created, it is critical to initialize the arra that it possesses (so that subsequent method calls on each Collector object can access/modif its arra without triggering a NullerEception). That is, here is a wrong implementation of the constructor of Collector: Collector() { /* wrong implementation of the constructor */ /* arra is implicitl initialized to */ } where the arra is not eplicitl initialized to certain length, so it is implicitl assigned to the default value of reference tpes:. Here is a visual summar of the effect of the above wrong implementation of constructor: Collector 0 Wh is it wrong? Tr access contents of the arra, for eample, calling pc.[0], then it will trigger a NullerEception, because pc. is and calling pc.[0] is equivalent to calling.[0], which simpl does not make sense. Line 5 initializes as an arra of size 100 (just for this eample as the assumed maimum number of that each Collectr ma collect). We onl initializes the arra with a fied size, without storing in an of its slots addresses of (simpl because initiall no have been collected et), so all slots in store values onl. Since there is no eplicit assignments that given an initial value to the attribute nop, it is initialized (implicitl) with the default value of integers: 0. Remark: The above digram shows a sensible configuration of objects: 1) pc..length is 100, meaning that we can in the future store at most 100 objects of tpe into it; and 2) pc.nop is 0, meaning that so far we have collected zero, and that the net (i.e., ver first) point object to be stored will be stored at pc.[0]. Mutator Method add of Collector : The mutator method void add(double, double ) (Lines 7 to 10, page 2) defines how a new point with coordinate values and can be added to the point collector. 7 void add(double, double ) { 8 [nop] = new (, ); /* store the new point at inde nop of */ 9 nop++; /* increment # of collected, update inde for storing net point */ 10 } There are two steps involved: Step 1 (Line 8): 8 [nop] = new (, ); Step 1.1: The RHS (i.e., new (, )) creates a new object, b calling the constructor of : the parameter values and of method add are used as argument values for the call to the constructor call (, ). Step 1.2: The LHS (i.e., [nop]) brings us to the specific slot at inde nop in arra. Step 1.: The middle equal sign (=) assigns/stores the address of the new object (created in Step 1.1) at inde nop in arra. Remember that one of the purposes of counter nop is that it indicates where a new to be collected should be stored. 4

5 Step 2 (Line 9): 9 nop++; Now that from Step 1 we alread store a new object at inde nop of arra, this line updates (b incrementing) the value of nop so that it still serves its purpose well. For eample, if in Step 1 the value of nop is 2, that means so far the slots [0], [1], [2] are all occupied b addresses of some object. So to make sure that value of nop correctl records the number of objects collected so far (which should be rather than 2), and that value of nop indicates the position for storing the net new to be collected (which should be [] rather than [2]), we increment the value of nop b 1. Accessor Method getsinquadranti of Collector : The accessor method [] getsinquadranti() (Lines 11 to 27, page 2) defines, out of the objects collected so far, how to return an arra containing onl those which are located at the first quadrant (i.e., those whose both coordinates are positive). First of all, we need to understand that simpl returning (which is also of tpe []) is not acceptable, because: 1) if nop <.length, there are.length - nop slots storing values; and 2) even if nop ==.length, not necessaril all collected so far are all located in the first quadrant. Therefore, this accessor method getsinquadranti is supposed to return an arra of objects, whose length corresponds to eactl the number of collected that are located in the first quadrant. There are three steps involved: Step 1 (Lines 12 to 20): 12 [] ps = new [nop]; 1 int count = 0; /* number of in Quadrant I (Q1) */ 14 for(int i = 0; i < nop; i ++) { 15 p = [i]; /* store the current point under consideration into p */ 16 if(p. > 0 && p. > 0) { /* the point being considered is in 1st quadrant */ 17 ps[count] = p; /* cop what s stored in p to slot at inde count of ps */ 18 count ++; /* increment # of Q1, update inde for storing net Q1 point */ 19 } 20 } Line 12 initializes an arra of size nop, indicating that might be up to nop that we have collected are located in the first quadrant. Wh? Given that so far we have collected nop (rather than.length) objects in the point collector, the maimum number of that are located in the first quadrant is simpl nop (i.e., when all collected are located in the first quadrant). As we iterate through the arra (slots [0], [1],..., [nop - 1]) some collected might be located in the first quadrant, and some might not be. As a result, we need a separate counter count that records eactl how man are located in the first quadrant. How count is used (Lines 17 and 18, Figure 2, page 2) to keep track of how man collected objects are located in the first quadrant is analogous to how nop is used (Lines 8 and 9, Figure 2, page 2) to keep track of how man objects are collected so far. Step 2 (Lines 21 to 25): 21 [] q1s = new [count]; 22 /* ps contains if count < nop */ 2 for(int i = 0; i < count; i ++) { 24 q1s[i] = ps[i] 25 } 5

6 From Step 1, we have updated the integer variable count so that it records the number of collected that are located in the first quadrant, and that the count slots ps[0], ps[1],..., ps[count - 1] store addresses of such first-quadrant. Notice that it is the case that count <= nop: Case 1) if count == nop, then all slots of arra ps (whose length is nop as can be seen in Line 12 in Figure 2, page 2) are occupied; Case 2) if count < nop, then at at least one slots of arra ps (i.e., ps[count], ps[count + 1],..., ps[nop]) store the (simpl because not all collected happen to be in the first quadrant). If the above Case 2 is true, meaning that there are values in arra ps, then we cannot simpl return ps as the return value for method getsinquadranti. Therefore: In Line 21 we initialize an arra q1s whose length corresponds the eact number of first-quadrant (i.e., the value of count). In Lines 2 to 25, we cop addresses of all such first-quadrant (stored in ps[0], ps[1],..., ps[count - 1]) to their corresponding positions in arra q1s to be returned: q1s[0], q1s[1],..., q1s[count - 1]. Step (Line 26): 26 return q1s; Now that the previous two steps store addresses of all first-quadrant (no more, and no less) into the arra q1s, we return it as the return value of method getsinquadranti (whose return tpe is []). 6

7 Class CollectorTester After understanding the descriptions of the attributes and methods of the classes (Section 1) and Collector (Section 2) from the previous two sections, let us learn how objects of these two classes ma interact. 1 class CollectorTester { 2 public static void main(string[] args) { Collector pc = new Collector(); 4 Sstem.out.println(pc.nop); /* 0 */ 5 pc.add(, 4); 6 Sstem.out.println(pc.nop); /* 1 */ 7 pc.add(-, 4); 8 Sstem.out.println(pc.nop); /* 2 */ 9 pc.add(-, -4); 10 Sstem.out.println(pc.nop); /* */ 11 pc.add(, -4); 12 Sstem.out.println(pc.nop); /* 4 */ 1 [] ps = pc.getsinquadranti(); 14 Sstem.out.println(ps.length); /* 1 */ 15 Sstem.out.println("(" + ps[0]. + ", " + ps[0]. + ")"); /* (, 4) */ 16 } 17 } Figure : Tester for Instances of and Tester Line of Tester : Collector pc = new Collector(); As eplained in the constructor of Collector (Section 2, page 2), the above line results in the following object structure: Object Structure After Initializing the Collector pc nop == 0 Collector Therefore, eecuting the print statement Sstem.out.println(pc.nop) in Line 4 of Collector outputs 0 to the console. 7

8 Line 5 of Tester : pc.add(, 4); This method call has the contet object pc, declared as an address holder of some Collector object, and the mutator method void add(double, double ) (defined in Lines 7 to 10 in Figure 2, Section 2, page 2) is instantiated (b substituting b and b 4) as: void add(, 4) { /* current value of nop is 0 */ [nop] = new (, 4); nop++; /* current value of nop is 1 */ } Eecuting the above method call results in the following object structure: Object Structure After Adding the 1st to the Collector pc nop == 1 Collector Therefore, eecuting the print statement Sstem.out.println(pc.nop) in Line 6 of Collector outputs 1 to the console. Line 7 of Tester : pc.add(-, 4); This method call has the contet object pc, declared as an address holder of some Collector object, and the mutator method void add(double, double ) (defined in Lines 7 to 10 in Figure 2, Section 2, page 2) is instantiated (b substituting b - and b 4) as: void add(, 4) { /* current value of nop is 1 */ [nop] = new (, 4); nop++; /* current value of nop is 2 */ } Eecuting the above method call results in the following object structure: 8

9 Object Structure After Adding the 2nd to the Collector pc nop == 2 Collector Therefore, eecuting the print statement Sstem.out.println(pc.nop) in Line 8 of Collector outputs 2 to the console. Line 9 of Tester : pc.add(-, -4); This method call has the contet object pc, declared as an address holder of some Collector object, and the mutator method void add(double, double ) (defined in Lines 7 to 10 in Figure 2, Section 2, page 2) is instantiated (b substituting b - and b -4) as: void add(, 4) { /* current value of nop is 2 */ [nop] = new (, 4); nop++; /* current value of nop is */ } Eecuting the above method call results in the following object structure: Object Structure After Adding the rd to the Collector pc nop == Collector Therefore, eecuting the print statement Sstem.out.println(pc.nop) in Line 10 of Collector outputs to the console. 9

10 Line 11 of Tester : pc.add(, -4); This method call has the contet object pc, declared as an address holder of some Collector object, and the mutator method void add(double, double ) (defined in Lines 7 to 10 in Figure 2, Section 2, page 2) is instantiated (b substituting b and b -4) as: void add(, 4) { /* current value of nop is */ [nop] = new (, 4); nop++; /* current value of nop is 4 */ } Eecuting the above method call results in the following object structure: Object Structure After Adding the 4th to the Collector pc nop == 4 Collector Therefore, eecuting the print statement Sstem.out.println(pc.nop) in Line 12 of Collector outputs 4 to the console. Pause and think: Up to this point, nop == 4 indicates that there have been four collected in the arra of the point collector object pc. We observe that out of these four collected, onl the stored at pc.[0] is located in the first quadrant. Therefore, we would epect the return value from a subsequent (accessor) method call pc.getsinquadranti() to be an arra of length 1. Line 1 of Tester : [] ps = pc.getsinquadranti(); This method call has the contet object pc, declared as an address holder of some Collector object, and the accessor method [] getsinquadranti() (defined in Lines 11 to 27 in Figure 2, Section 2, page 2) is eecuted (b substituting nop b its current value 4 and b the arra above where the first four slots are occupied) as: 10

11 Lines 12 to 20 of Collector : 12 [] ps = new [nop]; /* current value of nop is 4 */ 1 int count = 0; /* number of in Quadrant I */ 14 for(int i = 0; i < nop; i ++) { 15 p = [i]; 16 if(p. > 0 && p. > 0) { 17 ps[count] = p; 18 count ++; 19 } 20 } Right after Line 1, we initialize an arra ps of size nop (whose current value is 4, meaning that at most the four collected are all located in the first quadrant) and a counter count (whose job is to keep track of out of those collected so far, how man of them are located in the first quadrant). Here is a visual summar of the object structure after right after eecuting Line 1: nop == 4 Collector ps count == 0 Then, the for loop in Lines 14 to 20 is eecuted to iterate through pc.[0], pc.[1], pc.[2], and pc.[], and store whichever that is located in the first quadrant into slots in ps. Onl the first iteration (when i == 0) which inspects pc.[0] would pass the condition in Line 16, store the address of the point object (, 4) into ps, and update the count value accordingl; all later three iterations (when i is 1, 2, and ), the condition in Line 6 evaluates to false, so none of those three would be stored in ps and the value of count would not be incremented an further. Here is a visual summar of the object structure after right after eecuting Line 20: 11

12 nop == 4 Collector ps count == 1 It is important to notice that in Line 14, the value of nop (which is currentl 4), rather than the value of pc..length (which has remained 100 since it was first created), is used as the upper bound of the loop counter. What if i < pc..length was used instead as the sta condition of the for loop? Unfortunatel, when the value of loop counter i reaches 4 (which corresponds to the first, left-most inde at which pc. stores a value), we would still pass this (wrong) sta condition i < pc..length and enter the bod of the for loop. Consequentl, eecuting Line 15 assigns to variable p, and when the eecution reaches Line 16, eecuting p. and p. is equivalent to eecuting. and., which would trigger a NullerEception. Lines 21 to 25 of Collector : 21 [] q1s = new [count]; 22 /* ps contains if count < nop */ 2 for(int i = 0; i < count; i ++) { 24 q1s[i] = ps[i] 25 } From Lines 12 to 20, we have updated the values of count (which is now 1) and ps, so that arra ps now has 1 non- slot and (calculated through ps.length - count, where ps.length == nop && nop == 4) slots. We cannot simpl return arra ps as the return value for method getsinquadranti, as it contains slots. As a result, we eecute Lines 21 to 25 to initialize an arra q1s whose length corresponds to eactl those (indicated b the value of count) collected that are located in the first quadrant, and we simpl cop over the addresses of the first-quadrant to slots in arra q1s. 12

13 Figure 4 summarizes the object structure after right after eecuting Line 25: nop == 4 Collector q1s ps count == 1 Figure 4: Final Object Structure from Eecuting CollectorTester Again, it is important to notice that in Line 2, the value of count (which is currentl 1), rather than the value of nop (which is 4), is used as the upper bound of the loop counter. What if i < nop was used instead as the sta condition of the for loop? Unfortunatel, when the value of loop counter i reaches 1, we would still pass this (wrong) sta condition i < nop and enter the bod of the for loop. Consequentl, eecuting the epression q1s[i] in Line 24 would trigger an ArraIndeOutOfBoundEception because the arra q1s was created in Line 21 to have length count (which is 1). Line 26 of Collector : 26 return q1s; Up to now, there are three candidate objects (whose tpes all match the return tpe [] of the accessor method getsinquadranti) that can be the return values: 1. pc. This should not be returned, since it contains pc..length - nop slots that store values, and even for those slots that store addresses of objects, some of them are actuall not located in the first quadrant. 2. ps This should not be returned, since it contains nop - count slots that store value.. q1s This should be returned, since it contains eactl those collected that are located in the first quadrant, and there are no -slots in this arra. 1

14 Therefore, we return q1s (which stores the address of the arra of size 1) as the return value of accessor method getsinquadranti. This return value is stored, from Line 1 of CollectorTester (Figure, page 7), into variable ps. Consequentl: In Line 14 of CollectorTester, the print statement Sstem.out.println(ps.length) outputs 1 to the console. In Line 15 of CollectorTester, the print statement Sstem.out.println("(" + ps[0]. + ", " + ps[0]. + ")") outputs (, 4) to the console. 14

Implicit differentiation

Implicit differentiation Roberto s Notes on Differential Calculus Chapter 4: Basic differentiation rules Section 5 Implicit differentiation What ou need to know alread: Basic rules of differentiation, including the chain rule.

More information

Classes and Objects. EECS2030 B: Advanced Object Oriented Programming Fall 2018 CHEN-WEI WANG

Classes and Objects. EECS2030 B: Advanced Object Oriented Programming Fall 2018 CHEN-WEI WANG Classes and Objects EECS2030 B: Advanced Object Oriented Programming Fall 2018 CHEN-WEI WANG Separation of Concerns: App/Tester vs. Model In EECS1022: Model Component: One or More Java Classes e.g., Person

More information

Transformations of Functions. 1. Shifting, reflecting, and stretching graphs Symmetry of functions and equations

Transformations of Functions. 1. Shifting, reflecting, and stretching graphs Symmetry of functions and equations Chapter Transformations of Functions TOPICS.5.. Shifting, reflecting, and stretching graphs Smmetr of functions and equations TOPIC Horizontal Shifting/ Translation Horizontal Shifting/ Translation Shifting,

More information

Unit 5 Lesson 2 Investigation 1

Unit 5 Lesson 2 Investigation 1 Name: Investigation 1 Modeling Rigid Transformations CPMP-Tools Computer graphics enable designers to model two- and three-dimensional figures and to also easil manipulate those figures. For eample, interior

More information

Roberto s Notes on Differential Calculus Chapter 8: Graphical analysis Section 5. Graph sketching

Roberto s Notes on Differential Calculus Chapter 8: Graphical analysis Section 5. Graph sketching Roberto s Notes on Differential Calculus Chapter 8: Graphical analsis Section 5 Graph sketching What ou need to know alread: How to compute and interpret limits How to perform first and second derivative

More information

Announcements. CSCI 334: Principles of Programming Languages. Lecture 19: C++

Announcements. CSCI 334: Principles of Programming Languages. Lecture 19: C++ Announcements CSCI 4: Principles of Programming Languages Lecture 19: C++ HW8 pro tip: the HW7 solutions have a complete, correct implementation of the CPS version of bubble sort in SML. All ou need to

More information

class Point { int x; int y; Point(int x, int y) { this.x = x; this.y = y; void movevertically(int y){ this.y += y;

class Point { int x; int y; Point(int x, int y) { this.x = x; this.y = y; void movevertically(int y){ this.y += y; Call b Value (2.1) Aggregation and Comosition EECS200 B: Advanced Object Oriented Programming Fall 2018 CHEN-WEI WANG For illustration, let s assume the following variant of the class: class { int ; int

More information

Binghamton University. CS-140 Fall Problem Solving. Creating a class from scratch

Binghamton University. CS-140 Fall Problem Solving. Creating a class from scratch Problem Solving Creating a class from scratch 1 Recipe for Writing a Class 1. Write the class boilerplate stuff 2. Declare Fields 3. Write Creator(s) 4. Write accessor methods 5. Write mutator methods

More information

ACTIVITY 9 Continued Lesson 9-2

ACTIVITY 9 Continued Lesson 9-2 Continued Lesson 9- Lesson 9- PLAN Pacing: 1 class period Chunking the Lesson Eample A Eample B #1 #3 Lesson Practice M Notes Learning Targets: Graph on a coordinate plane the solutions of a linear inequalit

More information

Rational Functions with Removable Discontinuities

Rational Functions with Removable Discontinuities Rational Functions with Removable Discontinuities 1. a) Simplif the rational epression and state an values of where the epression is b) Using the simplified epression in part (a), predict the shape for

More information

6.1. Graphing Linear Inequalities in Two Variables. INVESTIGATE the Math. Reflecting

6.1. Graphing Linear Inequalities in Two Variables. INVESTIGATE the Math. Reflecting 6.1 Graphing Linear Inequalities in Two Variables YOU WILL NEED graphing technolog OR graph paper, ruler, and coloured pencils EXPLORE For which inequalities is (3, 1) a possible solution? How do ou know?

More information

Week 10. Topic 1 Polynomial Functions

Week 10. Topic 1 Polynomial Functions Week 10 Topic 1 Polnomial Functions 1 Week 10 Topic 1 Polnomial Functions Reading Polnomial functions result from adding power functions 1 together. Their graphs can be ver complicated, so the come up

More information

4.2 Properties of Rational Functions. 188 CHAPTER 4 Polynomial and Rational Functions. Are You Prepared? Answers

4.2 Properties of Rational Functions. 188 CHAPTER 4 Polynomial and Rational Functions. Are You Prepared? Answers 88 CHAPTER 4 Polnomial and Rational Functions 5. Obtain a graph of the function for the values of a, b, and c in the following table. Conjecture a relation between the degree of a polnomial and the number

More information

Optional: Building a processor from scratch

Optional: Building a processor from scratch Optional: Building a processor from scratch In this assignment we are going build a computer processor from the ground up, starting with transistors, and ending with a small but powerful processor. The

More information

Math 1050 Lab Activity: Graphing Transformations

Math 1050 Lab Activity: Graphing Transformations Math 00 Lab Activit: Graphing Transformations Name: We'll focus on quadratic functions to eplore graphing transformations. A quadratic function is a second degree polnomial function. There are two common

More information

science. In this course we investigate problems both algebraically and graphically.

science. In this course we investigate problems both algebraically and graphically. Section. Graphs. Graphs Much of algebra is concerned with solving equations. Man algebraic techniques have been developed to provide insights into various sorts of equations and those techniques are essential

More information

5.2 Graphing Polynomial Functions

5.2 Graphing Polynomial Functions Locker LESSON 5. Graphing Polnomial Functions Common Core Math Standards The student is epected to: F.IF.7c Graph polnomial functions, identifing zeros when suitable factorizations are available, and showing

More information

Pre-Algebra Notes Unit 8: Graphs and Functions

Pre-Algebra Notes Unit 8: Graphs and Functions Pre-Algebra Notes Unit 8: Graphs and Functions The Coordinate Plane A coordinate plane is formed b the intersection of a horizontal number line called the -ais and a vertical number line called the -ais.

More information

Functions: The domain and range

Functions: The domain and range Mathematics Learning Centre Functions: The domain and range Jackie Nicholas Jacquie Hargreaves Janet Hunter c 6 Universit of Sdne Mathematics Learning Centre, Universit of Sdne Functions In these notes

More information

Section 9.3: Functions and their Graphs

Section 9.3: Functions and their Graphs Section 9.: Functions and their Graphs Graphs provide a wa of displaing, interpreting, and analzing data in a visual format. In man problems, we will consider two variables. Therefore, we will need to

More information

CS 132 Compiler Construction, Fall 2011 Instructor: Jens Palsberg Multiple Choice Exam, Dec 6, 2011

CS 132 Compiler Construction, Fall 2011 Instructor: Jens Palsberg Multiple Choice Exam, Dec 6, 2011 CS 132 Compiler Construction, Fall 2011 Instructor: Jens Palsberg Multiple Choice Eam, Dec 6, 2011 ID Name This eam consists of 22 questions. Each question has four options, eactl one of which is correct,

More information

L3 Rigid Motion Transformations 3.1 Sequences of Transformations Per Date

L3 Rigid Motion Transformations 3.1 Sequences of Transformations Per Date 3.1 Sequences of Transformations Per Date Pre-Assessment Which of the following could represent a translation using the rule T (, ) = (, + 4), followed b a reflection over the given line? (The pre-image

More information

Enhanced Instructional Transition Guide

Enhanced Instructional Transition Guide Enhanced Instructional Transition Guide / Unit 04: Suggested Duration: 6 das Unit 04: Geometr: Coordinate Plane, Graphing Transformations, and Perspectives (9 das) Possible Lesson 0 (6 das) Possible Lesson

More information

6.867 Machine learning

6.867 Machine learning 6.867 Machine learning Final eam December 3, 24 Your name and MIT ID: J. D. (Optional) The grade ou would give to ourself + a brief justification. A... wh not? Problem 5 4.5 4 3.5 3 2.5 2.5 + () + (2)

More information

LINEAR PROGRAMMING. Straight line graphs LESSON

LINEAR PROGRAMMING. Straight line graphs LESSON LINEAR PROGRAMMING Traditionall we appl our knowledge of Linear Programming to help us solve real world problems (which is referred to as modelling). Linear Programming is often linked to the field of

More information

16.27 GNUPLOT: Display of functions and surfaces

16.27 GNUPLOT: Display of functions and surfaces 52 6.27 GNUPLOT: Displa of functions and surfaces This package is an interface to the popular GNUPLOT package. It allows ou to displa functions in 2D and surfaces in 3D on a variet of output devices including

More information

3.5 Rational Functions

3.5 Rational Functions 0 Chapter Polnomial and Rational Functions Rational Functions For a rational function, find the domain and graph the function, identifing all of the asmptotes Solve applied problems involving rational

More information

Exponential Functions. Christopher Thomas

Exponential Functions. Christopher Thomas Mathematics Learning Centre Eponential Functions Christopher Thomas c 1998 Universit of Sdne Mathematics Learning Centre, Universit of Sdne 1 1 Eponential Functions 1.1 The functions =2 and =2 From our

More information

Purely Functional Data structures (3) Purely Functional Data structures (4) Numerical Representations (1)

Purely Functional Data structures (3) Purely Functional Data structures (4) Numerical Representations (1) MGS 2011: FUN Lecture 2 Purel Functional Data Structures Henrik Nilsson Universit of Nottingham, UK Purel Functional Data structures (3) Linked list: After insert, if persistent: 1 2 3 1 7 Numerical Representations

More information

Pre-defined class JFrame. Object & Class an analogy

Pre-defined class JFrame. Object & Class an analogy CS1M Lecture 17 Mar 29, 25 1 Announcements: Project 4 due Sunda 4/3 at 6pm Use Keboard class or reading input Section in classrooms this week Previous Lecture: Selection statement Reading input using Keboard

More information

9 3 Rotations 9 4 Symmetry

9 3 Rotations 9 4 Symmetry h 9: Transformations 9 1 Translations 9 Reflections 9 3 Rotations 9 Smmetr 9 1 Translations: Focused Learning Target: I will be able to Identif Isometries. Find translation images of figures. Vocabular:

More information

1-1. Functions. Lesson 1-1. What You ll Learn. Active Vocabulary. Scan Lesson 1-1. Write two things that you already know about functions.

1-1. Functions. Lesson 1-1. What You ll Learn. Active Vocabulary. Scan Lesson 1-1. Write two things that you already know about functions. 1-1 Functions What You ll Learn Scan Lesson 1- Write two things that ou alread know about functions. Lesson 1-1 Active Vocabular New Vocabular Write the definition net to each term. domain dependent variable

More information

5.2 Graphing Polynomial Functions

5.2 Graphing Polynomial Functions Name Class Date 5.2 Graphing Polnomial Functions Essential Question: How do ou sketch the graph of a polnomial function in intercept form? Eplore 1 Investigating the End Behavior of the Graphs of Simple

More information

6.867 Machine learning

6.867 Machine learning 6.867 Machine learning Final eam December 3, 24 Your name and MIT ID: J. D. (Optional) The grade ou would give to ourself + a brief justification. A... wh not? Cite as: Tommi Jaakkola, course materials

More information

CS 403 Compiler Construction Lecture 8 Syntax Tree and Intermediate Code Generation [Based on Chapter 6 of Aho2] This Lecture

CS 403 Compiler Construction Lecture 8 Syntax Tree and Intermediate Code Generation [Based on Chapter 6 of Aho2] This Lecture CS 403 Compiler Construction Lecture 8 Snta Tree and Intermediate Code Generation [Based on Chapter 6 of Aho2] 1 This Lecture 2 1 Remember: Phases of a Compiler This lecture: Intermediate Code This lecture:

More information

INFORMATION CODING AND NEURAL COMPUTING

INFORMATION CODING AND NEURAL COMPUTING INFORATION CODING AND NEURAL COPUTING J. Pedro Neto 1, Hava T. Siegelmann 2, and J. Féli Costa 1 jpn@di.fc.ul.pt, iehava@ie.technion.ac.il, and fgc@di.fc.ul.pt 1 Faculdade de Ciências da Universidade de

More information

2.2 Absolute Value Functions

2.2 Absolute Value Functions . Absolute Value Functions 7. Absolute Value Functions There are a few was to describe what is meant b the absolute value of a real number. You ma have been taught that is the distance from the real number

More information

3x 4y 2. 3y 4. Math 65 Weekly Activity 1 (50 points) Name: Simplify the following expressions. Make sure to use the = symbol appropriately.

3x 4y 2. 3y 4. Math 65 Weekly Activity 1 (50 points) Name: Simplify the following expressions. Make sure to use the = symbol appropriately. Math 65 Weekl Activit 1 (50 points) Name: Simplif the following epressions. Make sure to use the = smbol appropriatel. Due (1) (a) - 4 (b) ( - ) 4 () 8 + 5 6 () 1 5 5 Evaluate the epressions when = - and

More information

Section 3.1: Introduction to Linear Equations in 2 Variables Section 3.2: Graphing by Plotting Points and Finding Intercepts

Section 3.1: Introduction to Linear Equations in 2 Variables Section 3.2: Graphing by Plotting Points and Finding Intercepts Remember to read the tetbook before attempting to do our homework. Section 3.1: Introduction to Linear Equations in 2 Variables Section 3.2: Graphing b Plotting Points and Finding Intercepts Rectangular

More information

2-1. The Language of Functions. Vocabulary

2-1. The Language of Functions. Vocabulary Chapter Lesson -1 BIG IDEA A function is a special tpe of relation that can be described b ordered pairs, graphs, written rules or algebraic rules such as equations. On pages 78 and 79, nine ordered pairs

More information

ECE 573 Midterm 1 September 29, 2009

ECE 573 Midterm 1 September 29, 2009 ECE 573 Midterm 1 September 29, 2009 Name: Purdue email: Please sign the following: I affirm that the answers given on this test are mine and mine alone. I did not receive help from an person or material

More information

CMSC 425: Lecture 10 Basics of Skeletal Animation and Kinematics

CMSC 425: Lecture 10 Basics of Skeletal Animation and Kinematics : Lecture Basics of Skeletal Animation and Kinematics Reading: Chapt of Gregor, Game Engine Architecture. The material on kinematics is a simplification of similar concepts developed in the field of robotics,

More information

Compiler construction

Compiler construction This lecture Compiler construction Lecture 5: Project etensions Magnus Mreen Spring 2018 Chalmers Universit o Technolog Gothenburg Universit Some project etensions: Arras Pointers and structures Object-oriented

More information

Graph Linear Equations

Graph Linear Equations Lesson 4. Objectives Graph linear equations. Identif the slope and -intercept of linear equations. Graphing Linear Equations Suppose a baker s cookie recipe calls for a miture of nuts, raisins, and dried

More information

ECE/CS 252: INTRODUCTION TO COMPUTER ENGINEERING UNIVERSITY OF WISCONSIN MADISON

ECE/CS 252: INTRODUCTION TO COMPUTER ENGINEERING UNIVERSITY OF WISCONSIN MADISON ECE/CS 252: INTRODUCTION TO COMPUTER ENGINEERING UNIVERSITY OF WISCONSIN MADISON Prof. Gurindar S. Sohi TAs: Newsha Ardalani and Rebecca Lam Midterm Eamination 3 In Class (5 minutes) Friday, November 8,

More information

Unit I - Chapter 3 Polynomial Functions 3.1 Characteristics of Polynomial Functions

Unit I - Chapter 3 Polynomial Functions 3.1 Characteristics of Polynomial Functions Math 3200 Unit I Ch 3 - Polnomial Functions 1 Unit I - Chapter 3 Polnomial Functions 3.1 Characteristics of Polnomial Functions Goal: To Understand some Basic Features of Polnomial functions: Continuous

More information

ACTIVITY: Graphing a Linear Equation. 2 x x + 1?

ACTIVITY: Graphing a Linear Equation. 2 x x + 1? . Graphing Linear Equations How can ou draw its graph? How can ou recognize a linear equation? ACTIVITY: Graphing a Linear Equation Work with a partner. a. Use the equation = + to complete the table. (Choose

More information

Wednesday, February 19, 2014

Wednesday, February 19, 2014 Wednesda, Februar 19, 2014 Topics for toda Solutions to HW #2 Topics for Eam #1 Chapter 6: Mapping High-level to assembl-level The Pep/8 run-time stack Stack-relative addressing (,s) SP manipulation Stack

More information

Online Homework Hints and Help Extra Practice

Online Homework Hints and Help Extra Practice Evaluate: Homework and Practice Use a graphing calculator to graph the polnomial function. Then use the graph to determine the function s domain, range, and end behavior. (Use interval notation for the

More information

SECTION 3-4 Rational Functions

SECTION 3-4 Rational Functions 20 3 Polnomial and Rational Functions 0. Shipping. A shipping bo is reinforced with steel bands in all three directions (see the figure). A total of 20. feet of steel tape is to be used, with 6 inches

More information

0 COORDINATE GEOMETRY

0 COORDINATE GEOMETRY 0 COORDINATE GEOMETRY Coordinate Geometr 0-1 Equations of Lines 0- Parallel and Perpendicular Lines 0- Intersecting Lines 0- Midpoints, Distance Formula, Segment Lengths 0- Equations of Circles 0-6 Problem

More information

A SCATTERED DATA INTERPOLANT FOR THE SOLUTION OF THREE-DIMENSIONAL PDES

A SCATTERED DATA INTERPOLANT FOR THE SOLUTION OF THREE-DIMENSIONAL PDES European Conference on Computational Fluid Dnamics ECCOMAS CFD 26 P. Wesseling, E. Oñate and J. Périau (Eds) c TU Delft, The Netherlands, 26 A SCATTERED DATA INTERPOLANT FOR THE SOLUTION OF THREE-DIMENSIONAL

More information

Polar Functions Polar coordinates

Polar Functions Polar coordinates 548 Chapter 1 Parametric, Vector, and Polar Functions 1. What ou ll learn about Polar Coordinates Polar Curves Slopes of Polar Curves Areas Enclosed b Polar Curves A Small Polar Galler... and wh Polar

More information

EECS2030 Fall 2016 Preparation Exercise for Lab Test 2: A Birthday Book

EECS2030 Fall 2016 Preparation Exercise for Lab Test 2: A Birthday Book EECS2030 Fall 2016 Preparation Exercise for Lab Test 2: A Birthday Book Chen-Wei Wang Contents 1 Before Getting Started 2 2 Task: Implementing Classes for Birthdays, Entries, and Books 3 2.1 Requirements

More information

Simple example. Analysis of programs with pointers. Program model. Points-to relation

Simple example. Analysis of programs with pointers. Program model. Points-to relation Simple eample Analsis of programs with pointers := 5 ptr := & *ptr := 9 := program S1 S2 S3 S4 What are the defs and uses of in this program? Problem: just looking at variable names will not give ou the

More information

Functions as Mappings from One Set to Another

Functions as Mappings from One Set to Another ACTIVITY. Functions as Mappings from One Set to Another As ou learned previousl, ordered pairs consist of an -coordinate and a -coordinate. You also learned that a series of ordered pairs on a coordinate

More information

Name: Thus, y-intercept is (0,40) (d) y-intercept: Set x = 0: Cover the x term with your finger: 2x + 6y = 240 Solve that equation: 6y = 24 y = 4

Name: Thus, y-intercept is (0,40) (d) y-intercept: Set x = 0: Cover the x term with your finger: 2x + 6y = 240 Solve that equation: 6y = 24 y = 4 Name: GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES SHOW ALL WORK AND JUSTIFY ALL ANSWERS. 1. We will graph linear inequalities first. Let us first consider 2 + 6 240 (a) First, we will graph the boundar

More information

Section 4.2 Graphing Lines

Section 4.2 Graphing Lines Section. Graphing Lines Objectives In this section, ou will learn to: To successfull complete this section, ou need to understand: Identif collinear points. The order of operations (1.) Graph the line

More information

CHECK Your Understanding

CHECK Your Understanding CHECK Your Understanding. State the domain and range of each relation. Then determine whether the relation is a function, and justif our answer.. a) e) 5(, ), (, 9), (, 7), (, 5), (, ) 5 5 f) 55. State

More information

Tester vs. Controller. Elementary Programming. Learning Outcomes. Compile Time vs. Run Time

Tester vs. Controller. Elementary Programming. Learning Outcomes. Compile Time vs. Run Time Tester vs. Controller Elementary Programming EECS1022: Programming for Mobile Computing Winter 2018 CHEN-WEI WANG For effective illustrations, code examples will mostly be written in the form of a tester

More information

Elementary Programming

Elementary Programming Elementary Programming EECS1022: Programming for Mobile Computing Winter 2018 CHEN-WEI WANG Learning Outcomes Learn ingredients of elementary programming: data types [numbers, characters, strings] literal

More information

Variables and Bindings

Variables and Bindings Net: Variables Variables and Bindings Q: How to use variables in ML? Q: How to assign to a variable? # let = 2+2;; val : int = 4 let = e;; Bind the value of epression e to the variable Variables and Bindings

More information

Section 2.2: Absolute Value Functions, from College Algebra: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a

Section 2.2: Absolute Value Functions, from College Algebra: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Section.: Absolute Value Functions, from College Algebra: Corrected Edition b Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution-NonCommercial-ShareAlike.0 license.

More information

3.6 Graphing Piecewise-Defined Functions and Shifting and Reflecting Graphs of Functions

3.6 Graphing Piecewise-Defined Functions and Shifting and Reflecting Graphs of Functions 76 CHAPTER Graphs and Functions Find the equation of each line. Write the equation in the form = a, = b, or = m + b. For Eercises through 7, write the equation in the form f = m + b.. Through (, 6) and

More information

Monday, September 28, 2015

Monday, September 28, 2015 Monda, September 28, 2015 Topics for toda Chapter 6: Mapping High-level to assembl-level The Pep/8 run-time stack (6.1) Stack-relative addressing (,s) SP manipulation Stack as scratch space Global variables

More information

Documenting, Using, and Testing Utility Classes

Documenting, Using, and Testing Utility Classes Documenting, Using, and Testing Utility Classes Readings: Chapter 2 of the Course Notes EECS2030: Advanced Object Oriented Programming Fall 2017 CHEN-WEI WANG Structure of Project: Packages and Classes

More information

Functions Project Core Precalculus Extra Credit Project

Functions Project Core Precalculus Extra Credit Project Name: Period: Date Due: 10/10/1 (for A das) and 10/11/1(for B das) Date Turned In: Functions Project Core Precalculus Etra Credit Project Instructions and Definitions: This project ma be used during the

More information

REMARKS. 8.2 Graphs of Quadratic Functions. A Graph of y = ax 2 + bx + c, where a > 0

REMARKS. 8.2 Graphs of Quadratic Functions. A Graph of y = ax 2 + bx + c, where a > 0 8. Graphs of Quadratic Functions In an earlier section, we have learned that the graph of the linear function = m + b, where the highest power of is 1, is a straight line. What would the shape of the graph

More information

1.5 LIMITS. The Limit of a Function

1.5 LIMITS. The Limit of a Function 60040_005.qd /5/05 :0 PM Page 49 SECTION.5 Limits 49.5 LIMITS Find its of functions graphicall and numericall. Use the properties of its to evaluate its of functions. Use different analtic techniques to

More information

4 B. 4 D. 4 F. 3. What are some common characteristics of the graphs of cubic and quartic polynomial functions?

4 B. 4 D. 4 F. 3. What are some common characteristics of the graphs of cubic and quartic polynomial functions? .1 Graphing Polnomial Functions COMMON CORE Learning Standards HSF-IF.B. HSF-IF.C.7c Essential Question What are some common characteristics of the graphs of cubic and quartic polnomial functions? A polnomial

More information

2.3. One-to-One and Inverse Functions. Introduction. Prerequisites. Learning Outcomes

2.3. One-to-One and Inverse Functions. Introduction. Prerequisites. Learning Outcomes One-to-One and Inverse Functions 2. Introduction In this Section we eamine more terminolog associated with functions. We eplain one-to-one and man-to-one functions and show how the rule associated with

More information

Transformations of Functions. Shifting Graphs. Similarly, you can obtain the graph of. g x x 2 2 f x 2. Vertical and Horizontal Shifts

Transformations of Functions. Shifting Graphs. Similarly, you can obtain the graph of. g x x 2 2 f x 2. Vertical and Horizontal Shifts 0_007.qd /7/05 : AM Page 7 7 Chapter Functions and Their Graphs.7 Transormations o Functions What ou should learn Use vertical and horizontal shits to sketch graphs o unctions. Use relections to sketch

More information

Chapter12. Coordinate geometry

Chapter12. Coordinate geometry Chapter1 Coordinate geometr Contents: A The Cartesian plane B Plotting points from a table of values C Linear relationships D Plotting graphs of linear equations E Horizontal and vertical lines F Points

More information

g(x) h(x) f (x) = Examples sin x +1 tan x!

g(x) h(x) f (x) = Examples sin x +1 tan x! Lecture 4-5A: An Introduction to Rational Functions A Rational Function f () is epressed as a fraction with a functiong() in the numerator and a function h() in the denominator. f () = g() h() Eamples

More information

Lesson 11 Skills Maintenance. Activity 1. Model. The addition problem is = 4. The subtraction problem is 5 9 = 4.

Lesson 11 Skills Maintenance. Activity 1. Model. The addition problem is = 4. The subtraction problem is 5 9 = 4. Lesson Skills Maintenance Lesson Planner Vocabular Development -coordinate -coordinate point of origin Skills Maintenance ddition and Subtraction of Positive and Negative Integers Problem Solving: We look

More information

Encapsulation in Java

Encapsulation in Java Encapsulation in Java EECS2030: Advanced Object Oriented Programming Fall 2017 CHEN-WEI WANG Encapsulation (1.1) Consider the following problem: A person has a name, a weight, and a height. A person s

More information

Pointers CMPE-013/L. Pointers. Pointers What do they do? Pointers What are pointers? Gabriel Hugh Elkaim Winter 2014

Pointers CMPE-013/L. Pointers. Pointers What do they do? Pointers What are pointers? Gabriel Hugh Elkaim Winter 2014 CMPE-013/L A Variable's versus A Variable's Value In some situations, we will want to work with a variable's address in memor, rather than the value it contains Gabriel Hugh Elkaim Winter 2014 Variable

More information

Lecture 15. Arrays (and For Loops)

Lecture 15. Arrays (and For Loops) Lecture 15 Arrays (and For Loops) For Loops for (initiating statement; conditional statement; next statement) // usually incremental { body statement(s); The for statement provides a compact way to iterate

More information

Recap. Recap. If-then-else expressions. If-then-else expressions. If-then-else expressions. If-then-else expressions

Recap. Recap. If-then-else expressions. If-then-else expressions. If-then-else expressions. If-then-else expressions Recap Epressions (Synta) Compile-time Static Eec-time Dynamic Types (Semantics) Recap Integers: +,-,* floats: +,-,* Booleans: =,

More information

Module Mechanisms CS412/413. Modules + abstract types. Abstract types. Multiple Implementations. How to type-check?

Module Mechanisms CS412/413. Modules + abstract types. Abstract types. Multiple Implementations. How to type-check? CS412/413 Introduction to Compilers and Translators Andrew Mers Cornell Universit Lecture 19: ADT mechanisms 10 March 00 Module Mechanisms Last time: modules, was to implement ADTs Module collection of

More information

Rotate. A bicycle wheel can rotate clockwise or counterclockwise. ACTIVITY: Three Basic Ways to Move Things

Rotate. A bicycle wheel can rotate clockwise or counterclockwise. ACTIVITY: Three Basic Ways to Move Things . Rotations object in a plane? What are the three basic was to move an Rotate A biccle wheel can rotate clockwise or counterclockwise. 0 0 0 9 9 9 8 8 8 7 6 7 6 7 6 ACTIVITY: Three Basic Was to Move Things

More information

CMSC 336: Type Systems for Programming Languages Lecture 4: Programming in the Lambda Calculus Acar & Ahmed 22 January 2008.

CMSC 336: Type Systems for Programming Languages Lecture 4: Programming in the Lambda Calculus Acar & Ahmed 22 January 2008. CMSC 336: Type Systems for Programming Languages Lecture 4: Programming in the Lambda Calculus Acar & Ahmed 22 January 2008 Contents 1 Announcements 1 2 Solution to the Eercise 1 3 Introduction 1 4 Multiple

More information

Four Ways to Represent a Function: We can describe a specific function in the following four ways: * verbally (by a description in words);

Four Ways to Represent a Function: We can describe a specific function in the following four ways: * verbally (by a description in words); MA19, Activit 23: What is a Function? (Section 3.1, pp. 214-22) Date: Toda s Goal: Assignments: Perhaps the most useful mathematical idea for modeling the real world is the concept of a function. We eplore

More information

Chapter 3. Interpolation. 3.1 Introduction

Chapter 3. Interpolation. 3.1 Introduction Chapter 3 Interpolation 3 Introduction One of the fundamental problems in Numerical Methods is the problem of interpolation, that is given a set of data points ( k, k ) for k =,, n, how do we find a function

More information

A Rational Shift in Behavior. Translating Rational Functions. LEARnIng goals

A Rational Shift in Behavior. Translating Rational Functions. LEARnIng goals . A Rational Shift in Behavior LEARnIng goals In this lesson, ou will: Analze rational functions with a constant added to the denominator. Compare rational functions in different forms. Identif vertical

More information

Search Trees. Chapter 11

Search Trees. Chapter 11 Search Trees Chapter 6 4 8 9 Outline Binar Search Trees AVL Trees Spla Trees Outline Binar Search Trees AVL Trees Spla Trees Binar Search Trees A binar search tree is a proper binar tree storing ke-value

More information

This exam is open book. Each question is worth 3 points.

This exam is open book. Each question is worth 3 points. This exam is open book. Each question is worth 3 points. Page 1 / 15 Page 2 / 15 Page 3 / 12 Page 4 / 18 Page 5 / 15 Page 6 / 9 Page 7 / 12 Page 8 / 6 Total / 100 (maximum is 102) 1. Are you in CS101 or

More information

1. A only. 2. B only. 3. both of them correct

1. A only. 2. B only. 3. both of them correct Version PREVIEW HW 10 hoffman (575) 1 This print-out should have 10 questions. Multiple-choice questions ma continue on the net column or page find all choices before answering. CalCe01b 001 10.0 points

More information

y = f(x) x (x, f(x)) f(x) g(x) = f(x) + 2 (x, g(x)) 0 (0, 1) 1 3 (0, 3) 2 (2, 3) 3 5 (2, 5) 4 (4, 3) 3 5 (4, 5) 5 (5, 5) 5 7 (5, 7)

y = f(x) x (x, f(x)) f(x) g(x) = f(x) + 2 (x, g(x)) 0 (0, 1) 1 3 (0, 3) 2 (2, 3) 3 5 (2, 5) 4 (4, 3) 3 5 (4, 5) 5 (5, 5) 5 7 (5, 7) 0 Relations and Functions.7 Transformations In this section, we stud how the graphs of functions change, or transform, when certain specialized modifications are made to their formulas. The transformations

More information

CS S-11 Memory Management 1

CS S-11 Memory Management 1 CS414-2017S-11 Management 1 11-0: Three places in memor that a program can store variables Call stack Heap Code segment 11-1: Eecutable Code Code Segment Static Storage Stack Heap 11-2: Three places in

More information

Using Characteristics of a Quadratic Function to Describe Its Graph. The graphs of quadratic functions can be described using key characteristics:

Using Characteristics of a Quadratic Function to Describe Its Graph. The graphs of quadratic functions can be described using key characteristics: Chapter Summar Ke Terms standard form of a quadratic function (.1) factored form of a quadratic function (.1) verte form of a quadratic function (.1) concavit of a parabola (.1) reference points (.) transformation

More information

Graphing Equations Case 1: The graph of x = a, where a is a constant, is a vertical line. Examples a) Graph: x = x

Graphing Equations Case 1: The graph of x = a, where a is a constant, is a vertical line. Examples a) Graph: x = x 06 CHAPTER Algebra. GRAPHING EQUATIONS AND INEQUALITIES Tetbook Reference Section 6. &6. CLAST OBJECTIVE Identif regions of the coordinate plane that correspond to specific conditions and vice-versa Graphing

More information

LESSON 3.1 INTRODUCTION TO GRAPHING

LESSON 3.1 INTRODUCTION TO GRAPHING LESSON 3.1 INTRODUCTION TO GRAPHING LESSON 3.1 INTRODUCTION TO GRAPHING 137 OVERVIEW Here s what ou ll learn in this lesson: Plotting Points a. The -plane b. The -ais and -ais c. The origin d. Ordered

More information

5. Minimizing Circuits

5. Minimizing Circuits 5. MINIMIZING CIRCUITS 46 5. Minimizing Circuits 5.. Minimizing Circuits. A circuit is minimized if it is a sum-of-products that uses the least number of products of literals and each product contains

More information

Classes. Compiling Methods. Code Generation for Objects. Implementing Objects. Methods. Fields

Classes. Compiling Methods. Code Generation for Objects. Implementing Objects. Methods. Fields Classes Implementing Objects Components fields/instance variables values differ from to usuall mutable methods values shared b all s of a class usuall immutable component visibilit: public/private/protected

More information

Using C# for Graphics and GUIs Handout #2

Using C# for Graphics and GUIs Handout #2 Using C# for Graphics and GUIs Handout #2 Learning Objectives: Review Math methods C# Arras Drawing Rectangles Getting height and width of window Coloring with FromArgb() Sierpinski Triangle 1 Math Methods

More information

Array Basics: Outline. Creating and Accessing Arrays. Creating and Accessing Arrays. Arrays (Savitch, Chapter 7)

Array Basics: Outline. Creating and Accessing Arrays. Creating and Accessing Arrays. Arrays (Savitch, Chapter 7) Array Basics: Outline Arrays (Savitch, Chapter 7) TOPICS Array Basics Arrays in Classes and Methods Programming with Arrays Searching and Sorting Arrays Multi-Dimensional Arrays Static Variables and Constants

More information

6-1: Solving Systems by Graphing

6-1: Solving Systems by Graphing 6-1: Solving Sstems b Graphing Objective: To solve sstems of linear equations b graphing Warm Up: Graph each equation using - and -intercepts. 1. 1. 4 8. 6 9 18 4. 5 10 5 sstem of linear equations: two

More information

Algebra I. Linear Equations. Slide 1 / 267 Slide 2 / 267. Slide 3 / 267. Slide 3 (Answer) / 267. Slide 4 / 267. Slide 5 / 267

Algebra I. Linear Equations. Slide 1 / 267 Slide 2 / 267. Slide 3 / 267. Slide 3 (Answer) / 267. Slide 4 / 267. Slide 5 / 267 Slide / 67 Slide / 67 lgebra I Graphing Linear Equations -- www.njctl.org Slide / 67 Table of ontents Slide () / 67 Table of ontents Linear Equations lick on the topic to go to that section Linear Equations

More information

PROBLEM SOLVING WITH EXPONENTIAL FUNCTIONS

PROBLEM SOLVING WITH EXPONENTIAL FUNCTIONS Topic 21: Problem solving with eponential functions 323 PROBLEM SOLVING WITH EXPONENTIAL FUNCTIONS Lesson 21.1 Finding function rules from graphs 21.1 OPENER 1. Plot the points from the table onto the

More information