# Chapter 3A - Rectangular Coordinate System

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1 - Chapter A Chapter A - Rectangular Coordinate Sstem Introduction: Rectangular Coordinate Sstem Although the use of rectangular coordinates in such geometric applications as surveing and planning has been practiced since ancient times, it was not until the 7th centur that geometr and algebra were joined to form the branch of mathematics called analtic geometr. French mathematician and philosopher Rene Descartes (96-60) devised a simple plan whereb two number lines were intersected at right angles with the position of a point in a plane determined b its distance from each of the lines. This sstem is called the rectangular coordinate sstem (or Cartesian coordinate sstem). -ais -ais origin (0, 0) Points are labeled with ordered pairs of real numbers,, called the coordinates of the point, which give the horizontal and vertical distance of the point from the origin, respectivel. The origin is the intersection of the -and-aes. Locations of the points in the plane are determined in relationship to this point 0,0. All points in the plane are located in one of four quadrants or on the -or-ais as illustrated below. To plot a point, start at the origin, proceed horizontall the distance and direction indicated b the -coordinate, then verticall the distance and direction indicated b the -coordinate. The resulting point is often labeled with its ordered pair coordinates and/or a capital letter. For eample, the point units to the right of the origin and units up could be labeled A,. Quadrant II Quadrant I (-, +) (+, +) (0, b) (a, 0) (0,0) Quadrant III Quadrant IV (-, -) (+, - ) Notice that the Cartesian plane has been divided into fourths. Each of these fourths is called a quadrant and the are numbered as indicated above.

2 - Chapter A Eample : Solution: Plot the following points on a rectangular coordinate sstem: A, B 0, C, D,0 E, C(-,) E(-,-) B(0,-) D(,0) A(,-) Eample : Shade the region of the coordinate plane that contains the set of ordered pairs, 0. [The set notation is read the set of all ordered pairs, such that 0.] Solution: This set describes all ordered pairs where the -coordinate is greater than 0. Plot several points that satisf the stated condition, e.g.,,, 7,,,0. These points are all located to the right of the -ais. To plot all such points we would shade all of Quadrants I and IV. We indicate that points on the -ais are not included 0 b using a dotted line. Eample : Shade the region of the coordinate plane that contains the set of ordered pairs,,. Solution: The area to the right of the dotted line designated is the set of all points where the -coordinate is greater than (shaded gra). The area between the horizontal lines designated and is the area where the -coordinate is between and (shaded red). The dark region is the intersection of these two sets of points, the set that satisfies both of the given conditions. = = = > and - < < The basis of analtic geometr lies in the connection between a set of ordered pairs and its graph on the Cartesian coordinate sstem.

3 - Chapter A Definitions: An set of ordered pairs is called a relation. The plot of ever point associated with an ordered pair in the relation is called the graph of the relation. The set of all first elements in the ordered pairs is called the domain of the relation. The set of all second elements in the ordered pairs is called the range of the relation. In Eample, we plotted five distinct points. If we consider these points as a set of ordered pairs, we have the relation,, 0,,,,,0,,. The graph is C(-,) A(,-) E(-,-) - - B(0,-) D(,0) The domain is, 0,,, and the range is,,, 0,. Infinite sets of ordered pairs can be described algebraicall and plotted (or graphed) on the coordinate sstem. Eample : Below is the graph from Eample. Recall that the graph represents all ordered pairs defined b the algebraic statement: 0. That is, the relation consists of all ordered pairs, that have an -coordinate that is a positive number. What is the domain and range of this relation? Solution: Since the relation is defined as the set of all ordered pairs where 0, the domain is 0. The -coordinates can be an real number so the range is all real numbers.

4 - Chapter A Eample : Below is the graph from Eample. What is the domain and range? = = = > and - < < Solution: The domain is all real numbers greater than. The range is all real numbers between and, including the endpoints and. Note: We often write the domain and range in interval notation. The domain for the above eample in interval notation is,. The range for the above eample in interval notation is,.

5 - Chapter A Distance Formula The marriage of algebra and geometr allows us to devise algebraic formulas to use in solving geometric problems. For eample, the formula for finding the distance between two points in the plane is derived as follows: Consider two points P, and Q,. Select a third point R, so that the three points form a right triangle with the right angle at point R. (See figure below.) Q(, ),, P( ) R( ) Note that the distance between P and R is and the distance from Q and R is. Therefore, the distance from point P to point Q, denoted d P, Q, can be found using the Pthagorean Theorem: d P, Q Because and, and because we are onl interested in positive values for the distance, we obtain the following formula. Distance Formula The distance between two points P, and Q, in the plane is d P, Q Eample : Find the distance between the points A, and B, 7. Solution: To find the distance between two points in the plane we use the distance formula d A,B. It does not matter which point ou use as, or,,so we will use the coordinates of A and B respectivel. That is,,,, and 7. Plugging into the formula, we get d A,B

6 - Chapter A Eample : triangle. Solution: Determine whether the points A,, B, and C 0, are the vertices of a right Plot and label points A, B, andc on graph paper and draw the indicated triangle. A(-,) C(0,) B(,) From inspection of the graph, point C appears to be the verte of the right triangle; therefore the line segment through A and B appears to be the hpotenuse. We will use the distance formula to find the length of each side of the triangle, and then appl the Pthagorean Theorem to determine whether the triangle is indeed a right triangle: If triangle ABC is a right triangle, d A,B d A,B d B,C 0 9 d A,C must be true. Since 7 9, triangle ABC is NOT a right triangle. Eample : Determine whether the points A, and B, are equidistant from point C,. Recall that equidistant means equal distance. That is, points A and B are equidistant from point C if and onl if the distance from A to C is equal to the distance from B to C. Algebraicall, we would write the above statement as d A,C d B,C. Solution: Plot the points on graph paper to visualize the problem B(-,-) - C(,) A(,) - Using the distance formula we can find the distance between each of the points, A and B, and point C. d A,C d B,C Because d A,C d B,C, A and B are equidistant from point C. Note that if ou plot all the points that are units from point C ou will obtain a circle.

7 - Chapter A Midpoint Formula In man instances it is important to be able to calculate the point that lies half wa between the two points on the line segment that connects them. The figure below shows points A, and B,, along with their midpoint M,. B(, ) = M(, ) Q(, ) = A(, ) P(, ) Note that Δ APM ΔBQM b ASA so that AP MQ. Solving for, we get. Similarl, Midpoint Formula. The midpoint of the line segment that connects points A, and B, is the point M, with coordinates,. The midpoint is the point M, with and. Note that the -coordinate of the midpoint is the average of the -coordinates of the the endpoints, and the -coordinate of the midpoint is the average of the -coordinates of the endpoints. The eamples below will show ou how to appl this formula. Be sure to work through them before tring the eercises. Eample : Find the midpoint of the line segment that connects the points, 9 and 7,. Solution: Plugging, 9, 7, and into the mid-point formula, we get 7 9 Therefore, the midpoint is,.

8 - Chapter A Eample : If, 8 is the midpoint of the line segment connecting A, and B, findthe coordinates of the other endpoint B. Solution: We are given the values, 8, and. We must find and. Substituting into the midpoint formula, we get and Solving the above for and, we get: Therefore, the endpoint B has coordinates, 8.

9 - Chapter A Circles The set of all points P, that are a given fied distance from a given fied point is called a circle. The fied distance, r, is called the radius and the fied point C h,k is called the center. r P(, ) C(h, k) The equation for this set of points can be found b appling the distance formula. That is, d C,P r. h k r If we square both sides of this last equation we get h k r Equation of a circle with C h, k and radius r Standard Form. h k r. Circles with center at the origin,c 0, 0 Standard Form r Eample : Find the center and radius or the circles a) 6 and b) 0. Solution: a) This circle is in the second form so its center is the origin C 0,0 and r 6 r b) This circle is in the first form so that h andk C, and r 0 r 0 Eample : Write an equation for the circle with C, and radius. Solution To write an equation for a specific circle we first write the equation for a circle in standard form: h k r, and then identif the specific values for h,k, andr.. Since we are given the center C and radius r, we can fill in values for h, k, andr as follows: h the -coordinate of C ; k the -coordinate of C ; and r. The equation of the circle is Substitute into the equation and simplif: Equation of the circle in standard form:

10 - Chapter A Eample : Find the center and radius of the circle 9. Graph the circle and find its domain and range. Solution: From the equation above, we see that h, k sothecenterisc,. Since r 9, the radius is r. Graph: Note that there are no points to the left of the point, nor to the right of,. Therefore the domain is all real numbers from to, including and, or the interval,. Similarl, the range values include all real numbers between 6 and 0, including the endpoints, which is the interval 6,0. Note:We can find the domain of the circle algebraicall b adding and subtracting the radius to the h value.,,. We can obtain the range b adding and subtracting the radius from the k value., 6,0. Eample : Write an equation for the circle with at the origin and radius. Solution: Substituting h 0, k 0, and r into the equation for a circle, we get 0 0 Since the center of the circle is the origin, we can substitute directl into the formula r to get. If we epand the equation for the circle, we get

11 - Chapter A Simplifing and arranging terms, we get We call this the general form of an equation of a circle. Notice that the coefficients of and are the same. This is our first clue that a particular equation ma be a circle. The equation h k r form is called the standard form of an equation of a circle. General Form of an Equation of a Circle c d e 0 It is important to be able to recognize that this equation also represents a circle. Note that the equation contains both an and a term and that both coeffiients equal. The general form is not as user-friendl as the standard form. We cannot find the center and radius of the circle b simpl inspecting the equation as we can with an equation in standard form. To find the center and radius of a circle that is in general form, we must reverse the above process and write the equation in standard form. For eample, 0 is the equation of a circle. (The coefficients of and are positive and equal.) Grouping the and terms and moving the constant to the other side of the equation, we get We must now complete the square on the and terms, and add the calculated amounts to both sides of the equations. Question: Answer: Wh did we add to both sides of the equation? Weaddedgobeabletowrite as a square.. So that we re not changing the equation we must either add and subtract from the same side of the equation 0, or we must add to the each side of the equation ( is equivalent to. Writing in factored form, we get the standard form of the equation: 9 From this form we can determine that the center of the circle is, and the radius is. We can use this information to graph the equation b plotting the center, and locating as man points on the circle as needed ( units in an direction from the center).

12 - Chapter A Eample : Find the center and radius of the circle 6 0. Graph the circle. Solution: To find the center and radius we must write the equation in standard form: Group and terms: 6 0 Move constant term to other side: 6 Complete the square: Rewrite in factored form: 9 Determine C and r: C, r Graph b plotting the center C, and appling the radius of units to find points on the circle: Since the radius is, the domain is, the range is,,. 0,6 and Eample 6: Find the center and radius of the circle 6 0. Solution: Although the coefficients of and are not, the equation represents a circle because the are equal, so we divide the equation b the common coeffient. 6 Divide equation b : 0 0 Group and terms: 0 Move constant term to other side: Complete the square: Rewrite in factored form: Determine C and r: C, r 9

13 - Chapter A Eercises for Chapter A - Rectangular Coordinate Sstem. Plot the following points on a rectangular coordinate sstem: a) A, b) B, c) C 0, d) D, e) E,0 f) F,. Use the points in eercise # to answer the following questions. a) Which point(s) are in quadrant I? quadrant II? quadrant III? quadrant IV? b) Which point(s) are on the -ais? the -ais? c) Which point(s) meet the condition: 0? d) Which point(s) meet the condition: 0? e) Which point(s) meet both the conditions: 0and?. Shade the region of the coordinate plane that contains each of the following sets of points. a), 0 b), and 0 c), and d), and. Graph the following relations and write their domain and range. a) 7,,,, 0,,, 6,,, 0,0, 9, b),,,,,.,.,,.6,.7. Write the domain and range of each of the relations in eercise. 6. Find the distance between the following sets of points. a) A, and B, b) C 0, and D, c) E,0 and F, d) G, e) J.,.9 and K., Is P 7, closer to point R, or point Q 9,? 8. Prove that the triangle with vertices A,, B,, andc 6,0 is a right triangle. 9. Find the area of triangle ABC in eercise #9. 0. Determine whether the triangle with vertices A,, B,, andc 9, is an isosceles triangle. (An isosceles triangle has two sides that are equal.). If the distance betweeen A, and B 6,0 is units, find all possible coordinates for A.. Write an equation that describes all the points, that are units from point B 6,0.. If the center of a circle is,9 and 0, is a point on the circle, find the radius of the circle.. Find the midpoint of the line segment connecting the following pairs of points: a) A, and B, b) C 0, and D, c) E,0 and F, d) G, and H, e) J.,.9 and K., 7.. In each of the following, M, is the mid-point of A and B. Find the missing end-point, A or B. a) A, ; M, b) M,0 ; B, 6. If 7, and,8 are endpoints of a diameter of a circle, find the center.

14 - Chapter A 7. Each of the following points A is an endpoint of a line segment. If the midpoint of the line segment AB is the point 0,0, findb. a), b), c), d), 6 e),0 8. Find the C h,k and radius r for the following circles. Then graph the circle. a) 7 6 b) 9 c) 7 d) 0 e) 8 0 f) Write an equation for each of the following circles. a) C, ; r b) C 0,0 ; r c) C, through 7, d) C 8, that touches the -ais at 0, 0. If,7 and, are endpoints of a diameter of a circle, write the equation for the circle.. Prove that the point C, is equidistant from A, and B 7,. Is C the midpoint of AC? Verif our answer.. If the diagonals (line segments connecting opposite vertices) of a parallelogram (a quadrilateral whose opposite sides are equal and parallel) are equal, the parallelogram is a rectangle (a parallelogram with four right angles). If all sides of the rectangle are equal, it is a square. Determine whether the quadrilateral with vertices A,, B,7, C,8, and D, is a parallelogram, rectangle, or square.

15 - Chapter A Answers to Eercises for Chapter A - Rectangular Coordinate Sstem. D(-,) A(-,-). a) QI: B; QII: D; QIII: A; QIV: F b) -ais: E; -ais: C c) B and F d) A, C, E, andf e) C and F. a) E(-,0) C(0,-) = 0 B(,) F(,-) b) = - = 0 c) = - = d)

16 - Chapter A = = = -. a) Domain: 7,,,,,9 Range:,,, 6,, b) Domain:,,,.,.6 Range:,,/,,.7. a) Domain: all real numbers 0 Interval notation:,0 Range: all real numbers Interval notation:, b) Domain: all real numbers greater than Interval notation:. Range: all real numbers 0 Interval notation: 0, c) Domain: all real numbers or, Range: all real numbers or, d) Domain: real numbers between and, including or, Range: real numbers or, 6. a) d A,B 6 6 Note: 6 is between 9 7 and 6 8. b) d C,D 0

17 - Chapter A. 669 c) d E,F d) d G,H. e) d J,K It s alwas helpful to graph the given situation on graph paper to help ou visualize the problem. Rememeber what formulas ou have to work with and what information each formula provides. To determine whether point R or point Q is closer to P, wemust determine the distance from R to P and from Q to P. The smaller of the two distances will tell us which point is closer. d P,R 6 d P,Q 0 Since d P,Q d P,R, Q is closer to P. 8. Plot the three points on graph paper, label them and draw the triangle. Remember that ou can prove that a triangle is a right triangle using the Pthagorean Theorem: a b c where a and b are the sides of the triangle and c is the hpotenuse. What formula we can use to find the length of each side? The distance formula, of course. d A,B 7 d B,C 68 7 d A,C 8 Once we have the lengths of each side, we plug the smaller sides into the Pthagoream Theorem for a and b, and the largest in for the hpotenuse Therefore, ΔABC is a right triangle. 9. To find the area of a triangle we can use the formula A bh, whereb is the base and h is the height. Note that h is the perpendicular distance from the third verte back to the base. Since ΔABC is a right triangle, the two legs are the base and the height, so that A TheareaofΔABC is 7square units. 0. Remember to plot the points and draw the triangle to help ou visualize the problem. To show that ΔABC is an isoscleles triangle we must show that two of the three sides are equal in length. Your plot above will probabl show ou which sides to tr first. d A,B 8 d B,C 8 Since sides AB and BC are equal in length ΔABC is an isosceles triangle.. Stating the problem algebraicall gives us d A,B. Replacing the left side with the distance formula, we get We must square both sides of the equation and solve for. 9 9

18 - Chapter A 9 6 Therefore A can be either of the points, and,. Justif this to ourself b plotting the points on graph paper.. d P,B Plot point B on graph paper. Sketch out the points that are units from point B. What shape do ou get? A circle.. Plot the center and point and sketch the circle. The radius of a circle r is the distance from the center of the circle to an point on the circle. Using the distance formula, we get r Find the midpoint of the line segment connecting the following pairs of points: a) Using the midpoint formula gives us M, M, b) M 0, M, c) M, M d) M,, M, e) M ,.,.6 M.0,.. a) Use Eample in the midpoint notes as a model for this problem. You should get the point 8, as our answer. b) 7, 6. Plot the given points on graph paper, draw the circle throught the two points, and draw the diameter. The diameter of a circle will alwas go through its center. Moreover, the center will be the mid-point of the diameter. To find the center, we must then find the midpoint of the two endpoints of the diameter using the midpoint formula. Center : 7, 8, 7. a), b), c), d),6 e),0 Note that the signs of the coordinates are opposites. If the origin is the midpoint of two points, we sa that the two points have smmetr with respect to the origin. 8. a) 7 6 Recall the formulas for circles. C,7 ; r b) C 0,0 ; r 7 c) C, ; r 7 d) C 0, ; r 0 e) Group -terms and move to other side of the equation: 8 Complete the square on the -terms b adding 8 6 to both sides of the equation Simplif: Therefore, C,0 ; r f) 0 0 0

19 - Chapter A C, ; r 0 9. a) b) c) We are given the center of the circle, so h andk, but we do not have the radius. Plot the center and draw the circle through P. What do we need to do to find the length of the radius? The distance formula! r d C,P 7 8 Substituting into the formula, we get 8 8 d) Plot the points on graph paper. You should be able to see that the radius of the circle is 8. Therefore the equation is The diameter of a circle goes through the center of a circle. Therefore, ou can use the midpoint formula to find the center of the circle. C, The radius is the distance from the center to either of the points on the circle. r. Plot out the problem on graph paper. Use the distance and midpoint formulas appropriatel to draw our conclusions algebraicall. You should find that C is equidistant from A and B, but it is not the midpoint. Eplain to ourself wh. Draw all the points that are equidistant from A and B. You should have drawn a line. This line is called the perpendicular bisector of line segment AB.. Using the distance formula appropriatel should show that all sides are equal in length and the diagonals are equal, therefore the most accurate term for the quadrilateral is square. Note that a square is also a parallelogram and a rectangle.

20 - Chapter B Chapter B - Graphs of Equations Graphing b Plotting Points We have seen that the coordinate sstem provides a method for locating points in a plane. Furthermore, we can plot sets of ordered pairs on the coordinate sstem to visualize the relationship between the two variables. However,in most cases we will be interested in relations that are stated as equations. For eample, the equation 6. refers to the relation, 6, read the set of all pairs, such that 6". Ever pair of numbers, that makes the equation true is called a solution to the equation. The pair, is a solution to the equation 6 because 6; but the ordered pair, is not a solution since 6. Question: Answer: How man solutions does the equation 6 have? There are an infinite number of solutions. Eample : Find three more pairs, such that 6. Solution: An pair of numbers whose sum is 6 is a solution. For eample:,,.,8.,,, 0,6, 6,0,,,, The plot of ever point that corresponds to a solution to the equation 6 is the graph of the equation 6: (-,) 0 (-.,8.) (.,.) (0,6) (,) (,) (6,0) The points labeled above are the solutions we listed in eample ; however, the graph consists of ever pair, that is a solution to the equation. Furthermore, ever point on the graph satisfies the equation 6.. Note that the above graph is a line. The graph of an linear equation is a line. A linear equation is one that can be written in the form a b c 0, where a and b are not both 0. The equation 6 can be written in the form of a linear equation 6 0, where a, b, and c 6. IMPORTANT CONCEPT: The graph of an equation consists of all pairs, that are solutions to the equation. Ever solution to the equation is a point on the graph and ever point on the graph, is a solution to the equation.

21 - Chapter B Eample : Which of the following pairs are points on the graph of 0? a),8 b), c), d) 0,0 e),9 Solution: a),8 is a solution because is a true statement. Therefore,8 is a point on the graph. b), is not a solution because Therefore,, is not on the graph. c), is not a solution because Therefore,, is not on the graph. d) 0,0 is a solution is a true statement. Therefore, 0,0 is a point on the graph. e),9 is a solution Therefore,,9 is a point on the graph. To find a solution to an equation, choose a value for, substitute it into the equation and solve for. Eample : Find three solutions to the equation 0. Solution: If, 0 0,0 is a solution. If, 0, is a solution. If, 0, is a solution. We can write this in chart form:, 0,0,, Note that each of the pairs, will be coordinates of points on the graph of 0.

22 - Chapter B Question: Is the point 0,0 on the graph of 0? Answer: 0,0 is not a point on the graph. For if it were, then the following equations would be true In fact, for the same reason, an point of the form 0, is not on the graph of the equation 0. Eample : Find fifteen solutions to the equation. Solution: Since we are going to substitute values for and solve for, it will be easier to first solve for : Substitute values for into, and solve for., 6 6 0,0,,,,,, 0 0 0,,,,,,,,,, 0,0 You should recognize that this equation gives us a graph that is a circle, and each of the points we found above is a point on the circle. Note that an values for that are less than or greater than will give us negative values under the radical sign. Therefore, -coordinates of the solutions will be in the interval,, which is the domain of the relation. To graph an equation b plotting points:. Solve the equation for.. Complete a table of values b substituting our choice of values for into the equation, then solving for. Use as man points as necessar to determine the shape of the graph.. Plot the points and draw a smooth curve through them..

23 - Chapter B Eample : Graph the equation. Solution: Solve for : Complete a table of values b picking an, and then using the above equation to determine the corresponding. Choose enough points to determine the shape of the graph. Plot the points and draw the curve., 6,6...,. 0,0 0...,. 0 0, 0...,. 0,0...,. 6, Note that the points on this graph do not form a straight line. This is because the equation is not in the form of a linear equation because is raised to the second power. Therefore, the graph will not be a line. This graph is called a parabola. The graph of a quadratic equation is a parabola. A quadratic equation is an equation that can be written in the form a b c, where a 0. The equation is a quadratic, where a, b 0, c. Note that an value for will give a real value for, so that the domain of the relation is,. The -values include all real numbers greater than or equal to, so that the range is,. Eample 6: Graph the equation. Solution: Solve for :. Complete a table of values. Plot the points and draw the curve.,, 0,0,,, 0 0, Note that the graph continues infinitel to the left and to the right, because we could place an number in the equation for and get a corresponding value for. Therefore, the domain of the relation is,. From the graph ou should be able to see that there are no -values larger than, so that the range is,

24 - Chapter B Eample 7: Graph the equation. Solution: Solve for :. Complete a table of values. Plot the points and draw the curve., 0 no solution no solution 0,0. 0.., 0.,, Note that the graph continues infinitel to the right because we could place an number in the equation for that is greater than and get a corresponding value for. Question: What happens to the graph when? Answer: When numbers were substituted for that were less than, we could not compute because we had a negative value under the radical. Therefore the graph does not etend to the left of the point,0. Since must be greater than or equal to for to be a real number, the domain of the relation is,. B eamining the graph and table of values, note that the -coordinates are greater than or equal to 0, so that the range is 0,

25 - Chapter B Intercepts Points that will prove to be ver important to our future stud of equations and their graphs are the intercepts. Definition: The points where the graph of an equation crosses the -ais is called the -intercept. If a,0 is a point on the graph of an equation, we sa that a is an -intercept of the graph because a,0 is a point on the -ais. Note that the -intercept will alwas have a -coordinate of 0. The -intercepts are also called zeros of the equation. A zero is a value for that causes the corresponding -value to be 0. If the point,0 is an -intercept for the graph of an equation, isazero of the equation. Definition: The points where the graph of an equation crosses the -ais is called the -intercept. If 0,b is a point on the graph of an equation, we sa that b is a -intercept of the graph because 0,b is a point on the -ais. Note that the -intercept will alwas have an -coordinate of 0. B inspecting the table of values and the graph of, we see the equation crosses the -ais at the point 0, and the -ais at the points,0 and,0. Therefore, isthe-intercept and and are-intercepts (,0) (0,-) Finding the -intercepts To find the -intercept(s) of an equation, substitute 0 for, and solve for. Finding the -intercepts To find the -intercept(s) of an equation, substitute 0 for, and solve for. Eample : Find the -and-intercepts of algebraicall. Solution: Find the intercepts b substituting 0 for the appropriate variable. -intercept: Substitute 0 for and solve for : 0 is the -intercept, and 0, is a point on the graph. -intercept: Substitute 0 for and solve for : and andare-intercepts and,0 and,0 are points on the graph.

26 - Chapter B When graphing b plotting points, it is important to include points close to andoneithersideofthe-intercepts. Question: Wh is the above true? Answer: The -intercepts are the zeros that is, the are the values for that cause 0. The -values ma change sign on either side of a zero which means that a graph ma be below the -ais on one side of an -intercept and above it on the other, or vice-versa. Numbers close to zero often behave differentl. For eample, if 0,. Prove this to ourself using several fractions that are between 0 and. E: Eample : Find the intercepts of the graph of. Solution: -intercepts: Substitute 0 for and solve for : 0 Thus, andare-intercepts; 0, and 0, are points on the graph. -intercepts: Substitute 0 for and solve for : 0 Thus, andare-intercepts;,0 and,0 are points on the graph. We can confirm our findings b inspecting the graph of which we know to be a circle with C 0,0 and radius. (-,0) 6 (0,) (0,-) (,0) Eample : Find the intercepts of 6. Solution: -intercepts: Substitute 0 for and solve for : 0, Thus, the point 0, 6 is the -intercept. -intercepts: Substitute 0 for and solve for :, and 0 6 and Thus, the two points 6,0 and,0 are the -intercepts.

27 - Chapter B Eample : Find the intercepts of 6 8. Solution: -intercepts: Substitute 0 for and solve for : 0, / Thus, the -intercept is the point 0,. -intercepts: Substitute 0 for and solve for :, and 0 and Thus, the -intercepts are the points,0 and.0.

28 - Chapter B Smmetr When we think of smmetr in art, we think of balance on both sides of a center of focus. In other situations, the term conjures up the idea of a mirror image. In either case, the notion of smmetr means essentiall the same thing in an algebra class: a shape looks the same on both sides of a dividing line or point. The dividing line is called the line of smmetr. We will also consider points of smmetr in this section. Eamine the graph of the circle for smmetr Question: How man lines or points of smmetr can ou find for the circle? Answer: The circle has smmetr about its center a point, and about an straight line that goes through its center an infinite number of lines. Although an line that goes through the center of the circle can be a line of smmetr for the circle, we will focus our efforts in this section on the -and-aes. We will also look at smmetr about the origin. Eample : A, B, andc. The figures on the aes below are smmetric about the -ais. Find the coordinates of Solution: A, B, 7 C, Note that the -coordinates are the same as the corresponding point in the Quadrant I figure, but the -coordinates are the negatives of the -coordinates in the Quadrant I figure (, ) A (, ) C (, 7) B Eample : Find the points of the figure that would be smmetric, about the -ais, to the above figure in Quadrant I. Solution: Corresponding points would be,,7,. Note that the -coordinates of corresponding points are the same as the Quadrant I figure, but the -coordinates are the negatives.

31 - Chapter B Test for Smmetr about the origin: Substitute for, and for into the equation. If the resulting equation is equivalent to the original, the graph has smmetr about the origin. Eample 6: Test for smmetr about the origin. Solution: Substitute for, and for, and simplif. Since the resulting equation is not equivalent to the original equation, there is no smmetr about the origin. Note that even if we multipl the resulting equation b, the result is not equivalent to the original equation. The results of the eamples in this section have shown that has smmetr about the -ais onl. Eample 7: Test the equation for smmetr about the -ais, -ais, and origin. Solution: -ais: [Substitute for ]: Since the resulting equation is equivalent to the original, the graph has smmetr about the -ais. -ais: [Substitute for ]:. equation has smmetr about the -ais. origin: [Substitute for and for ]: equation has smmetr about the origin. -

32 - Chapter B Eample 8: Test the equation for smmetr about the -ais, -ais, and origin. Solution: -ais: [Substitute for ]: equation does not have smmetr about the -ais. -ais: [Substitute for ]: equation does not have smmetr about the -ais. origin: [Substitute for and for ]: equation has smmetr about the origin Knowing whether the graph of a particular equation has smmetr about a point or line can assist us if we are graphing b plotting points. If we are using a grapher it provides us information about what our graph should look like so that we can determine whether a particular graph is reasonable for the equation we entered. Consider the section of graph in the figure below. If we knew that the graph had smmetr about the -ais, -ais, or origin, we could complete the graph knowing onl what the Quadrant I section of the graph looked like. Eample 9: Reproduce the graph above using paper and pencil. Complete the graph for each of the following situations. The graph has smmetr about the a) -ais b) -ais c)origin. Solution: (a) (b) (c)

33 - Chapter B To use smmetr in graphing an equation,. Determine whether the graph has smmetr about the -ais, -ais, or origin.. Plot enough points in the first quadrant to get an idea of what the graph looks like.. Appl smmetr to complete the graph. Eample 0: Use smmetr in graphing the equation. Solution: Determine smmetr. -ais: No smmetr about -ais. -ais: No smmetr about -ais. origin: Smmetr about the origin. Because has smmetr about the origin, use values of that are positive in the table of values, and use smmetr to find the remainder of the graph. / / Since, for eample, the point /, is on the graph of, and this graph is smmetric about the origin, we know that the point /, is also a point on the graph of. -

34 - Chapter B Eercises for Chapter B - Graphs of Equations. Graph. Graph 6.. Graph.. Graph.. Graph. 6. Graph. 7. Graph. 8. Find the intercepts for Find the intercepts for Find the intercepts for.. Find the intercepts for.. Find the intercepts for.. Find the intercepts for.. Test each equation for smmetr about the -ais, -ais, or origin. a) 0 b) 0 c) 9. d) 6 e) f). Use smmetr about the -ais, -ais, or origin to sketch the graph of each. a) b) c) d) 6. In one of the previous eamples we plotted some points on the graph of,and did not plot an points for which was positive. Find values of which satisf this equation for /0. Hint this equation is a quadratic in.

35 - Chapter B Answers to Eercises for Chapter B - Graphs of Equations (, ) (, -)

36 - Chapter B 6. (-, ) (, -) 7. (-, -) (, ) 8. -intercept: Substitute 0: or 0, -intercept: Substitute 0: or,0 9. -intercept: Substitute 0: or 0,6 -intercept: Substitute 0: 0 6 0, or,0,,0 0. -intercept: Substitute 0: 0 or 0, -intercept: Substitute 0: 0 or,0,,0. -intercept: Substitute 0: 0 or 0, -intercept: Substitute 0: no -intercepts. -intercept: Substitute 0: or 0, -intercept: Substitute 0: , or,0,,0. -intercept: Substitute 0: 0 no -intercepts -intercept: Substitute 0: 0 0 no -intercepts. a)-ais: Substitute for : 0 0 No smmetr -ais: Substitute for : 0 0 No smmetr Origin: Substitute for and for : 0 0 Multipling both sides b : 0 Smmetr Conclusion: Smmetr about the origin onl. b) -ais: Substitute for : 0 0 No smmetr -ais: Substitute for : 0 0 Smmetr Origin: Substitute for and for : 0 0 No smmetr Conclusion: Smmetr about the -ais onl. c) -ais: Substitute for : 9 9 No smmetr -ais: Substitute for : 9 9 No smmetr Origin: Substitute for and for : 9 9

37 - Chapter B Multipling both sides b : 9 Smmetr Conclusion: Smmetr about the origin onl. d) -ais: Substitute for : 6 6 Smmetr -ais: Substitute for : 6 6 No smmetr Origin: Substitute for and for : 6 6 No smmetr Conclusion: Smmetr about the -ais onl. e) -ais: Substitute for : No smmetr -ais: Substitute for : Smmetr Origin: Substitute for and for : No smmetr Conclusion: Smmetr about the -ais onl. f) Smmetr about the -ais, -ais, and origin. You prove it!. a) smmetr about the -ais onl (0, -) b) smmetr about the -ais,, c) Smmetr about the -ais, -ais, and origin. - - d) Smmetr about the origin. (-, ) (, -)

38 - Chapter B 6. If, then we have Setting 0 we have /0 0 Thus, /

39 - Chapter C Chapter C - Linear Equations in Two Variables Linear Equations in Two Variables In this section we will eamine equations in and where both variables are raised to the first power. Definition: An equation that can be written in the form A B C, wherea and B are not both 0, is called a linear equation. Question: According to the definition, which of the following are linear equations? a) b) 0 c) d) e) Answer: a) Yes. A, B, C. and are raised to the first power. b) Yes. 0 can be written as 0, so that A, B 0, C. A and B are not both 0, the variables are raised to the first power. c) Yes. can be written as. A, B, C. Both variables are raised to the first power. d) No. is raised to the second power. e) No. is raised to the second power. Linear equations can be graphed as an other equation b plotting points. Eample : Graph the linear equation. Solution: Solve for. Fill in a table of values, 7,7, 0 0,,, Plot the points and sketch (-,7) (-,) (0,) (,) (,-) Notice that all of the points were on a line. You might guess that this is wh these equations are called linear equations. The graph of an linear equation is a straight line. Because an two points determine a line, we can graph a linear equation b plotting at least two points. The intercepts are a good choice because the are usuall eas points to find. Question: What are the -and-intercepts of the above graph? Answer: -intercept: Substitute 0. 0 or 0, -intercept: Substitute 0. 0 or,0

40 - Chapter C Eample : Graph the linear equation 8. Solution: Solve for : 8 Make a table of values: Plot the points and draw the line:,, 0 0, 7,7 0,0 (-,) (?,0) - (,0) (,6) (0,) We can see from both the table and the graph that the -intercept is. Question: What is the -intercept? Answer: If 0, The -intercept is 8. Eample : Graph the linear equation. Solution: We recognize that is a linear equation. Make a table of values: Plot the points and draw the line:, 0 0, 0,0, (0,) (,) (-/, 0) Eample : Graph the linear equation 6 0. Solution: We recognize that 6 0 is a linear equation. Solving for :. Note that the equation does not contain a variable. The onl condition that eists for a point to be included as a solution to the equation is that must be. Make a table of values: Plot the points and draw the line:, 0 0,,, All points with -coordinate lie on the horizontal line units below the -ais.

41 - Chapter C In the following investigation ou ma use a graphing calculator but record the table of values and the graph of each equation on graph paper before graphing the net equation. As ou graph each equation, observe the results before graphing the net equation. If ou have a TI-8 or TI-8, ou can enter the equation using the Y buttononthetopleft of the displa. (You must solve the equation for first.) Set the viewing window using the WINDOW to the right of the Y button. The following viewing window gives a FRIENDLY WINDOW: min.7, ma.7. scl tells ou how man units ou want each tick mark to represent. min., ma. will give a SQUARE WINDOW. Otherwise, the range values can be manipulated to include the part of the graph ou want to see. You can multipl each of the above settings to get other friendl and square windows. If ou do not use a friendl window, ou will get large decimals when ou trace the graph using the TRACE button. Use right and left arrows to move the cursor over the graph. The and at the bottom of the screen will give the coordinates of the point highlighted b the cursor. The cursor ma skip over the integral -intercepts if the window is not a friendl window. Eample : Graph the following lines on the same coordinate sstem. a) Which number was different in each equation? What effect did this have on the graph? b) Which number was the same in each equation? What did each line have in common? Solution: a) The coefficient of was different in each equation. The coefficient of affected the steepness of the graph. b) The constant was in each equation. The all had a -intercept of. Eample 6: Graph the following lines on the same coordinate sstem. a) Which number was different in each equation? What effect did this have on the graph? b) What number was the same in each equation? What did each line have in common? Solution: a) The constant was different in each equation. The -intercepts were different in each line. b) The coefficient of was in each equation. Each line had the same steepness. Eample 7: Graph the following lines on the same coordinate sstem and compare with the lines in Eample. What is different between the equation of the lines in this eample and the lines in Eample? What effect did this difference have on the graphs? Solution: The coefficients of in Eample are positive, but the are negative in this eample. Graphs in Eample go uphill from left to right. Graphs in this eample go downhill from left to right. Conclusions: (Graphs of lines in the form m b. m, the coefficient of, affects the steepness of the line. The larger m, the steeper the line. When m 0, the line goes uphill from left to right. We sa the line is increasing. When m 0, the line goes downhill from left to right. We sa the line is decreasing. The number b is the -intercept. If 0, then m 0 b b

42 - Chapter C Slopes of Lines The idea of steepness has applications in man areas of our lives. When we build a house, we must determine the steepness of a roof called the pitch of the roof. Certainl if we are skiers or cclists, we are interested in the steepness of a ski slope or roadwa. In some cases it is important that we be able to measure steepness. For eample, the Americans with Disabilities Act requires that a wheel chair ramp rise no more than 6 inches for 6 feet of horizontal distance, as shown below. Ramp 6in.=.ft. 6 ft. In the above eample, we would call the vertical change the rise and the horizontal distance the run. We describe the steepness as the ratio of the rise to the run. Consider the line below that goes through the points A, and B 6, (,) (6,) 6 - = run (6,) } rise - = The rise can be calculated b subtracting the -coordinates of the given points and the run can be calculated b subtracting the -coordinates of the points. The ratio of the rise to the run can be simplified to. This ratio is called the slope of the line. Definition: The slope of a line, m, is the ratio of the change in to the change in. m rise change in Δ run change in Δ The following formula generalizes the process we used above to calculate the slope of an line if we are given two points on the line: P, and P,. The rise Δ and the run Δ can be calculated as follows: Δ and Δ. Definition: The slope of the line through points P, and P, is m. Eample : Plot each pair of points on graph paper and find the slope of the line: a), and, b),7 and,7 c), and,6 d), and, Solution: a) m 7 7 b) m c) m d) m 6 0 undefined

43 - Chapter C Eample : Use our results to complete each sentence. a) The line goes uphill from left to right when the slope is. b) The line goes downhill from left to right when the slope is. c) The line is horizontal when the slope is. d) The line is vertical when the slope is. Answers: a) positive b) negative c) 0 d) undefined Investigation of Slopes with a Grapher Enter the equation into our grapher. Use TABLE mode to answer the following questions. Question: How much does change ever time increases b unit? Answer: units. For eample when changes from 0 to, changes from to. Quesiton: How much does change ever time increases b units? Answer: changes 6 units. For eample, when changes from 0 to, changes from to 7, a total of 6 units. Question: How much does change ever time increases b units? Answer: 9 units. For eample, when changes from to, changes from 7 to 6, a total of 9 units. change in Question: Write each of the above answers as a ratio: change in. Are these ratios equivalent? Answer: 6 9. All these ratios are equivalent to. Quesiton: Compare the slope of the line and the coefficient of in the equation of the line. Answer: The slope of the line is. The coefficient of in the equation of the line is. Eample : Enter the equation into our grapher. a) Use two points from our table to calculate the slope of the line. Compare with the coefficient of in the equation. b) Tr several more eamples of lines in the form m b. What can ou conclude about the relationship between the coefficient of of a linear equation that has been solved for and the slope of the line? Solution: a) m which is also the coefficient of in the equation. b) The coefficient of is the slope of the line. Conclusions: The coefficient of is the slope of the line. If the slope is positive, the graph goes uphill from left to right. We sa the line is increasing. If the slope is negative, the graph goes downhill from left to right. We sa the line is decreasing.

44 - Chapter C Not onl can we find the slope of a line b solving its equation for, but the form m b is especiall hand in graphing lines. We must use the " " form to enter the equation into the grapher, but it also makes pencil and paper graphing easier, as well. To graph a line using the m b form, Since the -intercept is b, plot the point 0,b on the -ais. change in From the -intercept, appl the slope, m change in rise run to locate a second point on the line. Draw the line. Eample : Graph. Solution: The -intercept is, so we plot the point 0,. The slope is which means that decreases units when increases unit. We then count one unit to the right and units down from the -intercept to plot our second point, and draw the line m = run= rise = Since the slope of the line is negative, we epect the line to go downhill from left to right. We sa this line is decreasing. Eample : Find the slope and -intercept of the line 6. Then graph. Solution: We must first solve the equation for : 6 6. Therefore, the slope m and the -intercept b. We first plot the point 0,. From that point we use the slope to rise and run. Plot the second point and draw the line rise run (0, -) We epect the line to go uphill from left to right since the slope is positive. We sa this line is increasing.

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