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6.5 Slope intercept form for Inequalities:

Linear Inequality: is a linear equation with an inequality sign (< , ≤, >, ≥). 6.5 Slope intercept form for Inequalities:. Solution of an Inequality: is an ordered pair (x, y) that makes the inequality true. GOAL:.

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6.5 Slope intercept form for Inequalities:

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  1. Linear Inequality: is a linear equation with an inequality sign (< , ≤, >, ≥) 6.5 Slope intercept formfor Inequalities: Solution of an Inequality: is an ordered pair (x, y) that makes the inequality true.

  2. GOAL:

  3. Whenever we are given a graph we must be able to provide the equation of the function. Slope-Intercept Form: The linear equation of a nonvertical line with an inequality sign: y (<, ≤, >, ≥) m x + b Slope = = y-intercepty crossing http://mathgraph.idwvogt.com/examples.html

  4. Whenever we are given a graph we must be able to provide the equation of the function. dash line shade left or down y < mx +b

  5. Whenever we are given a graph we must be able to provide the equation of the function. dash line shade right or up y > mx +b

  6. Whenever we are given a graph we must be able to provide the equation of the function. Solid line shade left or down y≤mx +b

  7. Whenever we are given a graph we must be able to provide the equation of the function. Solid line shade Right or up y ≥ mx +b

  8. EX: Provide the equation of the inequality.

  9. Solution: Since line is dashed and shaded at the bottom we use <.Also, the inequality must be in slope-intercept form: Y < mx + b A(0,1) 1. Find the y-intercept B(3,-2) In this graph b = +1. 2. Find another point to get the slope. A(0,5) B(3,-2)

  10. Use the equation of slope to find the slope: = = = -1 A(0,1) The slope-intercept form inequality is:y <-1x + 1 B(3,-2) Remember:This means that if you start a 1 and move down one and over to the right one, and continue this pattern. We shade the bottom since it is <.

  11. When work does not need to be shown: (EOC Test) look at the triangle made by the two points. Count the number of square going up or down and to the right. In this case 1down and1right. Thus slope is -1/1 = -1

  12. YOU TRY ITProvide the equation of the inequality.

  13. YOU TRY IT: (Solution)The inequality is solid and shaded below: Y ≤ mx + b 1. Find the y-intercept A(0,4) In this graph b = + 4. B(1,0) 2. Find another point to get the slope. A(0,4) B(1,0)

  14. Use the equation of slope to find the slope: = = = - 4 A(0,4) The slope-intercept form equation is:y ≤-4x + 4 B(1,0) Remember:This means that if you start a 4 and move down four and one over to the right. Solid line and shaded down means we must use ≤.

  15. When no work is required, you can use the rise/run of a right triangle between the two points: A(0,4) Look at the triangle, down 4 (-4) over to the right 1 (+1) slope = -4/+1 = -4 B(1,0) Remember:You MUST KNOW BOTH procedures, the slope formula and the triangle.

  16. Given Two Points: We can also create an inequality in the slope-intercept form from any two points and the words: less than (<), less than or equal to (≤), greater than(>), greater than and equal to(≥) accordingly. EX: Write the slope-intercept form of the line that is greater than or equal to and inequality that passes through the points (0, -0.5) and(2, -5.5)

  17. Use the given points and equation of slope: B(2,-5.5) A(0,-0.5) = = = - We now use the slope and a point to find the y intercept (b). y ≥ mx + b -3 = - + b -3 += b Isolate b: b =- = -

  18. Going back to the equation: y = mx + b we replace what we have found: m = - and b = - ½ To get the final slope-intercept form of the line passing through (3, -2) and(1, -3) y ≥ x – ½

  19. We now proceed to graph the equation: y ≥ -x - Y-intercepty crossing

  20. YOU TRY IT:Write the equation of the inequality.

  21. Use the given points and equation of slope: B(1,-3) A(3,-2) = = = We now use the slope and a point to find the y intercept (b). y < mx + b -3 = + b -3 -= b Isolate b: b = - = -

  22. Going back to the equation: y = mx + b we replace what we have found: m = and b = - To get the final slope-intercept form of the line passing through (3, -2) and(1, -3) y < x -3.5

  23. We now proceed to graph the equation: y < x - Y-intercepty crossing 2 1

  24. Real-World: A fish market charges $9 per pound for cod and $12 per pound per flounder. Let x = pounds of cod and y = pounds of flounder. What is the inequality that shows how much of each type of fish the store must sell per day to reach a daily quota of at least $120?

  25. Real-World(SOLUTION): A fish market charges $9 per pound for cod and $12 per pound per flounder. Let x = pounds of cod and y = pounds of flounder. What is the inequality that shows how much of each type of fish the store must sell per day to reach a daily quota of at least $120? Cod  x Flounder  y At least $120  9x + 12y ≥ 120

  26. SOLUTION: y ≥ - x + 10 9x + 12y ≥ 120  10 9 8 10 8 4 8 10 12 4 Any point in the line or in the shaded region is a solution.

  27. YOU TRY IT: A music store sells used CDs for $5 and buys used CDs for $1.50. You go to the store with $20 and some CDs to sell. You want to have at least $10 left when you leave the store. Write and graph an inequality to show how many CDs you could buy and sell.

  28. Real-World(SOLUTION): A music store sells used CDs for $5 and buys used CDs for $1.50. You go to the store with $20 and some CDs to sell. You want to have at least $10 left when you leave the store. Write and graph an inequality to show how many CDs you could buy and sell. Bought CDs  -5x Sold CDs  +1.5y At least $10 left  -5x + 1.5y ≥ -10 NOTE: -10 since you spent this money.

  29. SOLUTION: y ≥x – 6.6 -5x + 1.5y ≥ -10  10 8 6 4 2 2 3 4 1 Any point in the line or in the shaded region is a solution.

  30. VIDEOS: LinearInequalities https://www.khanacademy.org/math/algebra/linear-equations-and-inequalitie/graphing-linear-inequalities/v/graphing-inequalities https://www.khanacademy.org/math/algebra/linear-equations-and-inequalitie/graphing-linear-inequalities/v/solving-and-graphing-linear-inequalities-in-two-variables-1

  31. CLASSWORK:Page 393-395 Problems: As many as needed to master the concept.

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