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C ollege A lgebra

C ollege A lgebra. Functions and Graphs (Chapter1). Objectives. Cover the topics in Section ( 1-1):Cartesian Coordinate System. 1.1 : Cartesian Coordinate System. 1.2 : Graphing point by point. 1.3 : Distance between two points. 1.4 : Circles. Chapter1. y -axis. Quadrant II.

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C ollege A lgebra

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  1. College Algebra Functions and Graphs (Chapter1)

  2. Objectives Cover the topics in Section ( 1-1):Cartesian Coordinate System 1.1 : Cartesian Coordinate System. 1.2 : Graphing point by point. 1.3 : Distance between two points. 1.4 : Circles. Chapter1

  3. y-axis Quadrant II Quadrant I x-axis origin Quadrant III Quadrant IV 1.1 : Cartesian Coordinate System Rectangular ( Cartesian) Coordinate System The following is the rectangular coordinate system: It is made up of two number lines: The horizontal number line is the x- axis. 1.The vertical number line is the y- axis. 2.The origin is where the two intersect.  This is where both number lines are 0. It is split into four quadrants which are marked on this graph with Roman numerals. Chapter1

  4. 1.1 : Cartesian Coordinate System Rectangular ( Cartesian) Coordinate System Each point on the graph is associated with an ordered pair.  When dealing with an x, y graph, x is always first and y is always second in the ordered pair (x, y).  It is a solution to an equation in two variables.  Even though there are two values in the ordered pair, be careful that it associates to ONLY ONE point on the graph,  the point lines up with both the x value of the ordered pair (x-axis) and the y value of the ordered pair (y-axis). Chapter1

  5. 4 units 2 units 1.1 : Cartesian Coordinate System Plotting Points • Each point in the xy-plane corresponds to a unique ordered pair (a, b). • Plot the point (2, 4). • - Move 2 units right • - Move 4 units up Chapter1

  6. 1.1 : Cartesian Coordinate System Example 1:Plot the ordered pairs and name the quadrant or axis in which the point lies. A(2, 3), B(-1, 2), C(-3, -4), D(2, 0), and E(0, 5). A(2, 3) lies in quadrant I. B(-1, 2) lies in quadrant II. C(-3, -4) lies in quadrant III. D(2, 0) lies on the x-axis. E(0, 5) lies on the y-axis. Chapter1

  7. 1.1 : Cartesian Coordinate System Solutions of equations in 2 variables The solutions to equations in two variables consist of two values that when substituted into their corresponding variables in the equation, make a true statement.  In other words, if your equation has two variables x and y, and you plug in a value for x and its corresponding value for y and the mathematical statement comes out to be true, then the x and y value that you plugged in would together be a solution to the equation.  Equations in two variables can have more than one solution. We usually write the solutions to equations in two variables in ordered pairs.  Chapter1

  8. 1.1 : Cartesian Coordinate System Example 2:  Determine whether each ordered pair is a solution of the given equation.     y = 5x - 7;  (2, 3), (1, 5) Let’s start with the ordered pair (2, 3). Which number is the x value and which one is the y value?  If you said x = 2 and y = 3, you are correct!  Let’s plug (2, 3) into the equation and see what we get: This is a TRUE statement, so (2, 3) is a solution to the equation y = 5x - 7. Chapter1

  9. 1.1 : Cartesian Coordinate System Now let’s take a look at (1, 5).  Which number is the x value and which one is the y value?    If you said x = 1 and y = 5, you are right!  Let’s plug (1, 5) into the equation and see what we get: Whoops, it looks like we have ourselves a FALSE statement. This means that (1, 5) is NOT a solution to the equation 5x - 7. Chapter1

  10. Let x = 1 y = 5(1) + 2 y = 7 (1, 7) Let x = 0 y = 5(0) + 2 y = 2 (0, 2) Let y = 3 3 = 5x + 2 5 = 5x 1 = x (1, 3) 1.1 : Cartesian Coordinate System Example 3:  For the following equation find three ordered pairs that are solutions. y = 5x+ 2 Chapter1

  11. 1.2 : Graphing point by point Graphing a Linear Equation by Plotting Points The graph of an equation is found by plotting points that are solutions of the equation. The intercepts of the graph are good points to find first. x-intercept is an x-value where the graph intersects the x-axis. y = 0 y-intercept is a y-value where the graph intersects the y-axis. x = 0 11 Chapter1

  12. 1.2 : Graphing point by point Graphing a Linear Equation by Plotting Points Step 1:  Find three ordered pair solutions. You do this by plugging in ANY three values for x and find their corresponding y values. Yes, it can be ANY three values you want, 1, -3, or even 10,000.  Remember there are an infinite number of solutions.  As long as you find the corresponding y value that goes with each x, you have a solution. Step 2:  Plot the points found in step 1. Remember that each ordered pair corresponds to only one point on the graph. The point lines up with both the x value of the ordered pair (x-axis) and the y value of the ordered pair (y-axis). Step 3:  Draw the graph. A linear equation will graph as a straight line.  If you know it is a linear equation and your points don’t line up, then you either need to check your math in step 1 and/or that you plotted all the points found correctly. Chapter1

  13. 1.2 : Graphing point by point Example 4:  graph the equation.    y = 5x - 3 Step 1:  Find three ordered pair solutions. Step 2:  Plot the points found in step 1. Chapter1

  14. 1.2 : Graphing point by point Step 3:  Draw the graph. Chapter1

  15. 1.2 : Graphing point by point Graphing a Non-Linear Equation by Plotting Points If the equation is non-linear: Step 1:  Find six or seven ordered pair solutions. Non-linear equations can vary on what the graph looks like.  So it is good to have a lot of points so you can get the right shape of the graph. Step 2:  Plot the points found in step 1. Step 3:  Draw the graph. If the points line up draw a straight line through them.  If the points are in a curve, draw a curve through them. Chapter1

  16. 1.2 : Graphing point by point Example 5:  graph the equation.    Step 1:  Find six or seven ordered pair solutions Chapter1

  17. 1.2 : Graphing point by point Step 2:  Plot the points found in step 1. Step 3:  Draw the graph. Chapter1

  18. 1.3 : Distance between two points. The Distance Formula 18 Chapter1

  19. 1.3 : Distance between two points. The Distance Formula 19 Chapter1

  20. 1.3 : Distance between two points. The Midpoint Formula 20 Chapter1

  21. 1.3 : Distance between two points. The Midpoint Formula 21 Chapter1

  22. 1.4 : Circles. 22 Chapter1

  23. 1.4 : Circles. 23 Chapter1

  24. 1.4 : Circles. Example 1: Write the standard form of the equation of the circle with center (5, 7) andr=4 Putting it into standard form we get:        . Example 2: Write the standard form of the equation of the circle with center (-3, -1) and . Putting it into standard form we get: 24 Chapter1

  25. Step 1a:  Make sure that the coefficients on theandterms are equal to 1. If the coefficient ofand arealready 1, then proceed to step 1b. If the coefficient is not equal to 1, then divide both sides by the coefficients of theandsquared terms. 1.4 : Circles. The General Form of the Equation of a Circle Finding the Center and Radius of a Circle Given its Equation Step 1: Write the equation of the circle in standard form 25 Chapter1

  26. 1.4 : Circles. Step 1b:  Isolate the   ,x,    ,   andyterms. At this point we will be creating a perfect square trinomial for thexterms as well as theyterms.  Recall that a perfect square trinomial (PST) is of the formand it factors in the form          .  When it is in that form it will allow us to continue onto the next step and write the equation in standard form We need to find a number that we can add to the  x2     and x terms so that we have a PST.  We also need to find a number that we can add to the  y2     and y terms so that we have a PST.  We can get that magic number by doing the following: If we have we complete its square by addingto both sides of the equation. 26 Chapter1

  27. 1.4 : Circles. Step 1d:  Factor both  PST’s created in step 1c as a binomial squared. Step 2:Identify the center and radius of circle. 27 Chapter1

  28. 1.4 : Circles. Example 3:Find the center and radius of the circle  , and graph it. Step 1: Write the equation of the circle in standard form (                                ). The equation is already in standard form. The center is (3, -9) and the radius is 3. Step 3: Graph, if needed. 28 Chapter1

  29. 1.4 : Circles. Example 4:Find the center and radius of the circle   and graph it. Step 1: Write the equation of the circle in standard form (                                ). Step 1a:  Make sure that the coefficients on the  X2 and Y2 terms is equal to 1.   The coefficients on the  and  terms are both already 1. Step 1b:  Isolate the X2, x ,  Y2 ,and y terms. Step 1c:  Complete the square for bothxandy. 29 Chapter1

  30. 1.4 : Circles. Step 1d:  Factor both  PST’s created in step 1c as a binomial squared. The center is (-2, 1) and the radius is 4. Step 3: Graph, if needed. 30 Chapter1

  31. 1.4 : Circles. Example 5:Find the center and radius of the circle   and graph it. Step 1: Write the equation of the circle in standard form (                                ). Step 1a:  Make sure that the coefficients on the  X2 and Y2 terms is equal to 1.   The coefficients on the  and  terms are both already 1. Step 1b:  Isolate the X2, x ,  Y2 ,and y terms. Step 1c:  Complete the square for bothxandy. 31 Chapter1

  32. 1.4 : Circles. Step 1d:  Factor both  PST’s created in step 1c as a binomial squared. The center is (0, -3) and the radius is 1. Step 3: Graph, if needed. 32 Chapter1

  33. End of the Lecture Let Learning Continue

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