2.1-2.4 Linear Equations and Graphs Coordinate Plane Quadrants - The x-axis and y-axis form 4 "areas" known as quadrants. 1. I - The first quadrant has positive x and positive y points. 2. II - The second quadrant has negative x and positive y points. 3. III - The third quadrant has negative x and negative y points. 4. IV - The fourth quadrant has positive x and negative y points. Slope of a Line * The number of units a line rises or falls vertically for each unit of horizontal change. * m is used to represent slope in linear equations. Other forms of slope: Directions of slope: Positive slope m > 0 Negative slope m < 0 Zero slope m = 0 No slope/undefined Slope Division by 0 HOMEWORK EXAMPLE Exercises 5-8 on pg. 74
Finding Slope To find the slope between two points: 1) Use one of the slope formulas 2) Graph the points and count RISE over RUN. Directions: Find the slope between the points (4, 4) and (7, -2). Solution: (Numerically) Use one of the following formulas: (y 1 - y 2 ) / (x 1 - x 2 ) or (y 2 - y 1 ) / (x 2 - x 1 ) (4 - (-2)) / (4-7) or (-2-4) / (7-4) 6/-3 or -6/3-2 or -2 Solution: (Graphically) Plot the 2 points on graph paper and use Rise over Run: You can start from (4, 4) and go to (7, -2) or You can start from (7, -2)and go to (4, 4). You would get 6 / 3 or You would get 6 / 3-6/+3 or +6/-3-2 or -2 Directions: Verbally explain how you would find the slope between the points (5, 2) and (3, 2). Then find the slope. HOMEWORK EXAMPLE Exercise 15 on pg. 74
Equations of a Line Equations that are of degree 1 will be linear when represented by a graph. Forms for Equations of a line: Point-Slope Form: y - y 1 = m ( x - x 1 ) Used to find equations when give a point and the slope Slope-Intercept Form: y = mx + b Used to sketch a graph easily given an equation or the slope and the y-intercept Standard Form: Ax + By + C = 0 Mainly used to write equations when there is undefined slope Finding the Equation of a Line To Find an Equation (Algebraically) given slope and/or point(s): NOTE: To write an equation you will always need AT LEAST ONE POINT and SLOPE. 1) Find the slope 2) Use one of the points and plug into POINT-SLOPE Form To Find an Equation (Graphically) given slope and/or point(s): NOTE: If the graph doesn't pass through a point on the y-axis, this method will not work. 1) Find the slope 2) Use the slope to graph other points until you cross y-axis (this is your y-intercept) 3) Plug in your slope and y-intercept into SLOPE-INTERCEPT Form Directions: Find the equation of the line that passes through the points (4, 4) and (7, -2). Solution: (Algebraically) 1) Find the slope from the 2 points:. m = 4 - (-2) = 6 = -2 4-7 -3 2) Use POINT-SLOPE form: We can use either point: (4, 4) or (7, -2) Solution: (Graphically) 1) Plot the 2 points on graph paper 2) Use the slope you found in 1 of Finding Slope, m = -2/1, to find the y-intercept. 3) The y-intercept is 12. The point (0, 12). So b is 12. 4) We now have slope (m) and the y-intercept (b). So, we should use the Slope-Intercept Form. y = -2x + 12 NOTE: If your graph does not cross at an exact point on the y-axis, you will not be able to find b. HOMEWORK EXAMPLE Exercise 33 on pg. 92
Applications Using Linear Equations During 2000, Nike's net sales were $9.0 billion, and in 2001 net sales were $9.5 billion. Write a linear equation giving the net sales y in terms of the year x. Then use the equation to find the net sales for 2002. NOTE: Since they told us to use x and y, we will. Usually, when you are given a problem choose variables that make sense to you. Solution: (Algebraically) 1) Find "points" that we know. Our coordinates should look like this (year x, net sales y) According to the information given to us by Nike, our points should be (2000, 9) and (2001, 9.5) 2) Find the slope of the two points. 3) Put it into an equation and solve for y. Since we know the slope and we know a point, we should use Point-Slope Form. 4) Now we need to find the net sales (y) for 2002 (x). Since our time corresponds to x, we will plug 2002 in for x and solve for y (the net sales). 5) Always put your answer in the terms you were given in the problem. y = $10 billion HOMEWORK EXAMPLE Exercise 29 on pg. 91
Equations and their Graphs Methods to Graph Linear Equations: X-Y Table Method or the Plug-in Method 1) Use the X-Y Table Method by plugging in values for x and seeing what your corresponding y value is. 2) Then use those coordinate points to graph the line. : Graph y = -2x + 3 x y Corresponding Point -2 7 (-2, 7) 0 4 (0, 4) 2 1 (2, 1) NOTE: You should always use at least 3 points. Slope-Intercept Form 1) The equation must be in y = mx + b form 2) Find b. This is the y-intercept or your "starting point" on the y-axis. 3) Find m. This is your slope to get your next 2 points. : Graph y = -2x + 3 b = 3. So, we mark where 3 is on the y-axis. This is our first point. m = -2. So, we can use our slope to get our next 2 points. From our starting point, we go Down 2 units and Right 1 unit. This is our second point. To get our next point, we go from 2nd point and again go Down 2 units and Right 1 unit. NOTE: You could also go back to the starting point to get your 3rd point by going Up 2 units and Left 1 unit. HOMEWORK EXAMPLE Exercise 37 on pg. 82
Parallel and Perpendicular Lines Parallel Lines - Lines that have the same slope are parallel. Perpendicular Lines - Lines whose slopes are "negative reciprocal" of each other will be perpendicular. Finding Parallel Lines and Perpendicular Lines: 1) Find the Slope-Intercept form of the equation. 2) Find your slope 3) Transform slope so it is parallel/perpendicular 4) Plug in "new slope" and the parallel/perpendicular point. Find the the equation of the line that passes through the point (2, -1) and is parallel to the line 2x - 3y = 5 Solution: 1) Get the line into slope-intercept form: 2x - 3y = 5 (we'll call this our Original Line) 2) Find Slope: slope (m) for our Original Line is 2/3. 3) Transform slope: Since we want to find another line (We'll call this our Parallel Line) that is parallel to Original Line, our slope for our Parallel LIne must also be 2/3. 4) Plug in "new slope" and parallel/perpendicular point: We know it has m=2/3 and that it passes through the point (2, -1). So, we'll use the Point-Slope form to find our equation.
Find the equation of the line that passes through the point (2, -1) and is perpendicular to the line 2x - 3y = 5 Solution: 1) Get the line into slope-intercept form: 2x - 3y = 5 (we'll call this our Original Line) 2) Find Slope: slope (m) for our Original Line is 2/3. 3) Transform slope: Since we want to find another line (We'll call this our Perpendicular Line) that is perpendicular to Original Line, our slope for our Perpendicular LIne must be -3/2. 4) Plug in "new slope" and parallel/perpendicular point: We know it has m= -3/2 and that it passes through the point (2, -1). So, we'll use the Point-Slope form to find our equation. HOMEWORK EXAMPLE Exercise 51 on pg. 92