1. GRAPHING LINEAR EQUATIONS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice Levasseur

2. GRAPHING LINEAR EQUATIONS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice Levasseur

3. Basic Definitions  Axes – perpendicular number lines x-axis – horizontal number line y-axis – vertical number line Origin – the point of intersection of the axes  Ordered pair – a number pair (x, y), coordinate, point Abscissa – first coordinate, x Ordinate – second coordinate, y

4. Quadrant IQuadrant II Quadrant III Quadrant IV

5. Graphs On a number line, each point is the graph of a number On a plane, each point is the graph of a number pair ordered pair: (x, y) 02 (1,2)

6. Ex: Plot (3, 5) Ex: Plot (-3, 4) Ex: Plot (-4, -2)

7. An ordered pair is a solution to a 2-variable equation if a true statement results when the equation is evaluated at the ordered pair. Ex: Show (4, 5) and (-2, 2) are solutions to y = ½ x + 3 y = ½ x + 3 5½ (4) + 3 52 + 3 55 True  (4, 5) is a solution! y = ½ x + 3 2½ (-2) + 3 2-1 + 3 22 True  (-2, 2) is a solution!

8. Ex: Plot (4, 5) and (-2, 2) on the same set of axes. Notice a straight line connect the points (solutions)

9.  Connecting all the points that are solutions to an equation will result in a straight line.  In other words, the graph of the line connecting the points (solutions) represents the solution set of the equation.  Since the graph of the solution set is represented by a straight line, the equation is called a linear equation.  More formally, an equation in 2 variables, where the exponents of the variables are 1, is called a linear equation. Ax + By = C or y = mx + b, where A, B, C, m, and b are constants and A and B are not both 0

10. Graphing a Linear Equation (Plotting Points) 1.Solve the equation for one of the variables, usually y 2.Pick a value for x, plug it in, & solve for y 3.Repeat at least two more times 4.Plot the points on the same set of axes 5.Connect the dots with a straight line

11. Ex: Graph 6x – 3y = 3 Solve for a variable: y - 3y = 3 – 6x y = - 1 + 2x y = 2x - 1 x y 0 1 ½ y = 2(0) - 1 y = 2(1) - 11 y = 2( ½ ) - 1 0 We have identified 3 solutions to the equation: (0, -1)(1, 1)( ½, 0) 6x – 3y = 3 -6x -3

12. 1 1 Plotting the three solutions/points we get: (0, -1)(1, 1)( ½, 0) The solution points lie on a straight line. Every point on this line is a solution to the equation 6x – 3y = 3!

13. Your turn to try a few

14. We can always plot points to graph linear equations However, plotting points could be tedious, (especially for “messy” equations) There must be other ways to plot linear equations...

15. Graphing using Intercepts Consider a linear equation of the form Ax + By = C The y-intercept is the point in which the graph of the line crosses the y-axis, (0, b)  To find the y-intercept, let x = 0 and solve for y The x-intercept is the point in which the graph of the line crosses the x-axis, (a, 0)  To find the x-intercept, let y = 0 and solve for x

16. 1 1 Ex: Graph 2x – 3y = 6 using intercepts y-int: let x = 0 2(0) – 3y = 6 – 3y = 6 y = - 2 (0, - 2) x-int: let y = 0 2x – 3(0) = 6 2x = 6 x = 3(3, 0)

17. Your turn to try a few

18. Slope-Intercept Form Consider the linear equation Ax + By = C Solving for y, we get an equation of the form y = mx + b, where m and b are constants y = mx + b is called the slope-intercept form because b is the “intercept” (y-intercept (0, b)) and m is the “slope”

19. When the slope is negative (m < 0), the line slants down from left to right When the slope is positive (m > 0), the line slants up from left to right

20. Graphing using the slope and y-intercept Given the slope-intercept form, we can identify the slope, m, and the y-intercept, (0, b) To graph an equation of a line, given the slope- intercept form, start by plotting the y-intercept Then use the slope to identify at least 2 more solutions of the equation (i.e. solution points)

21. Recall, y = mx + b, where m and b are numbers, is the slope-intercept form of a linear equation. Ex: Find the slope-intercept form of x + 5y = 10 To find the slope-intercept form, we need to solve for y x + 5y = 10 5y = - x + 10 y = (-1/5) x + 2 y = m x + b m = - 1/5 b = 2 (0, 2) is the y-int 1 -x 5

22. Graph using Slope-Intercept form: y = (-1/5) x + 2 and y-int = (0, 2) 5 Slope m = Plot (0, 2) 2 2 Next, use the sloperise = - 1  down 1 run = 5  right 5 m = 1/(-5)  up 1, left 5 Note: -a/b = a/(-b) 

23. Ex: Graph 2x + 3y = -9 solve for y to use the slope-int form 2x + 3y = - 9 3y = - 2x - 9 y = (-2/3)x – 3 y = m x + b m = Negative slope  line slants down from left to right b = - 3  (0, -3) is the y-int -2 3  START HERE rise = -2  down 2run = 3  right 3 -2x 3

24. Graph: y = (-2/3)x - 3 and y-int = (0, -3) -2 3 Slope m =Plot (0, -3) 2 2 Next, sloperise = -2  down 2run = 3  right 3 orrise = 2  up 2run = -3  left 3

25. Your turn to try a few

26. Special Lines The graph of y = b is a horizontal line with y-intercept (0, b)  y is always b no matter what x is The graph of x = a is a vertical line with x-intercept (a, 0)  x is always a no matter what y is

27. 3 3 Ex: Graph 7x + 63 = 0 7x + 63 = 0 7x = - 63 x = - 9 x is always – 9 no matter what y is  vertical

28. 3 3 Ex: Graph 12y = 48 12y = 48 y = 4 y is always 4 no matter what x is  horizontal

29. Horizontal lines have slope m = 0 Vertical lines have undefined slope, m = undefined No vertical change  rise = 0  m = 0 No horizontal change  run = 0  m undefined m = rise run

30. To Graph a Linear Equation: Plot points: pick nice x and solve for y  (x, y), find at least 3 solutions Intercepts: oy-int: set x = 0, solve for y  (0, y) ox-int: set y = 0, solve for x  (x, 0) Slope-Intercept: y = mx + b oStart with y-int (0, b) oUse slope m to plot at least 2 more points Only one variable: oy = b  horizontal line with y-int (0, b) ox = a  vertical line with x-int (a, 0)