Comprehensive Guide to Solving Quadratic Inequalities in Two Variables

Comprehensive Guide to Solving Quadratic Inequalities in Two Variables

• 2.
  Math 10, Term1, Week 6
• 3.
  LEARNIN G COMPETE NCY Solve quadratic inequalities intwo variables.
• 4.
  DAY 01
• 5.
  SHORT REVIEW
• 6.
  Solve for xin 𝑥2 3 > 4 − 𝑥 1) Write the inequality in standard form. 2) Solve for the critical points. 3) Plot the critical points on a number line. 4) Make the sign table test. 5) Write the solution set. 𝑥2 3 -4 > 0 (+) − 𝑥 (x – 4)(x + 1) x = 4, x = -1 x < -1 –1 < x < 4 x > 4 Factors Test: x = –3 Test: x = 2 Test: x = 5 (x - 4) - - + (x + 1) - + + (x - 4)(x + 1) + - + True/False True False True x < -1 or x > 4 OR (– , -1) U (4, + ). ∞ ∞
• 7.
  LESSON PURPOSE
• 8.
  Warm-up Activity “WHAT CANYOU OBSERVE?”
• 9.
  “WHAT CAN YOUOBSERVE?” 1. Are all the graphs the same? If not, what are their differences?
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  “WHAT CAN YOUOBSERVE?” 2. Why are some graphs shaded while others are not? What does the shaded part represent?
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  “WHAT CAN YOUOBSERVE?” 3. Why do all of the graphs have a “U-shaped” form?
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  “WHAT CAN YOUOBSERVE?” 4. What is the difference between the broken lines and solid lines used in sketching the graph?
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  “WHAT CAN YOUOBSERVE?” 5. What are your other observations about the graphs?
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  “WHAT CAN YOUOBSERVE?” all of the graphs presented are quadratic except for the fact that only Graph 1 is an equation and the rest are inequalities
• 15.
  UNLOCKING CONTENT VOCABULARY
• 16.
  A quadratic inequalityin two variables describes a region of the Cartesian Plane with parabola as the boundary. A parabola is a U-shaped curve that is drawn for quadratic equations and quadratic
• 17.
  𝑦 < 𝑎𝑥2 +𝑏𝑥 + 𝑐 𝑦 > 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑦 ≥ 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑦 ≤ 𝑎𝑥2 + 𝑏𝑥 + 𝑐 Quadratic Inequalities in two variables can be written in any of the following forms, where a, b, and c are real numbers, and a ≠ 0. FORMS
• 18.
  𝑦 < 2𝑥2 +5𝑥 − 6 𝑦 > 𝑥2 2 − 𝑥 − 3 𝑦 ≥ 𝑥2 − 𝑥 − 12 𝑦 ≤ 2𝑥2 + 3 + 𝑥 1 Quadratic Inequalities in two variables can be written in any of the following forms, where a, b, and c are real numbers, and a ≠ 0. EXAMPLES
• 19.
  SOLVING QUADRATIC INEQUALITIES INTWO VARIABLES BY GRAPHING SUB- TOPIC 1
• 20.
  EXPLICI TATION
• 21.
  Let’s go backto these graphs…
• 22.
  “WHAT CAN YOUOBSERVE?” Graph 1 Quadratic Equation Graph 2 Quadratic Inequality Graph 3 Quadratic Inequality
• 23.
  = vs >vs ≥ y = x2 – 2x -1 y x ≥ 2 – 2x -1 y > x2 4x −
• 24.
  A dash orbroken line is used for inequalities with < or > symbols to describe that the points on the parabola are not solutions. On the other hand, a solid line is used for inequalities with or ≤ ≥ symbols to indicate that the points on the parabola are solutions.
• 25.
  The Parabola opens upwardif a > 0 or a is positive Example: y > x2 – 4x a = + The Parabola opens downward if a < 0 or a is negative Example: y -x ≤ 2 – 2x + 3 a = -1
• 26.
  VERTEX The turning point. Also,the highest or lowest point. If the given is an inequality, use the formula Vertex = (h,k) =
• 27.
  DAY 02
• 28.
  SHORT REVIEW
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  TRUE OR FALSE -4≤ 1 FALSE TRUE
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  TRUE OR FALSE -2› -3 FALSE TRUE
• 31.
  TRUE OR FALSE 0< -1 FALSE TRUE
• 32.
  TRUE OR FALSE 5≥ -5 FALSE TRUE
• 33.
  TRUE OR FALSE 6≥ -7 FALSE TRUE
• 34.
  WORKED EXAMPLE
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  STEP 1: Rewritethe inequality as a quadratic equation where the right side is 0. 𝑥2 2 3 = 0 − 𝑥 − STEP 2: Solve the equation by factoring, quadratic formula, or completing the square as the case may be and find its roots. 𝑥2 2 3 = 0 − 𝑥 − ( 3)( + 1) = 0 𝑥 − 𝑥 𝑥 = 3, = 1 𝑥 − The roots of the equation are (3, 0) and (-1, 0). 𝑦 < 𝑥2 2 3 − 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution.
• 36.
  STEP 3: Findthe vertex of the equation. 𝑦 < 𝑥2 2 3 − 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution. Vertex = (h, k) = = = = = The vertex is (1, -4).
• 37.
  The roots ofthe equation are (3, 0) and (-1, 0). The vertex is (1, -4). 𝑦 < 𝑥2 2 3 − 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution.
• 38.
  STEP 4: Selecta point ( , ) in each 𝑥 𝑦 region (inside and outside the parabola) and check whether the given inequality is satisfied 𝑦 < 𝑥2 2 3 − 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution. Test point (0, 0) which is found inside the parabola. 𝑦 < 𝑥2 2 3 − 𝑥 − 0 < 02 2(0) 3 − − 0 < 0 0 3 − − 0 < 3 − 𝐹𝐴𝐿𝑆𝐸 Test point (-2, 1) which is found outside the parabola. 𝑦 < 𝑥2 2 3 − 𝑥 − 1 < ( 2) − 2 2( 2) 3 − − − 1 < 4 + 4 3 − 1 < 5 𝑇𝑅𝑈𝐸
• 39.
  STEP 5: Shadethe region that contains the solution. Shade the region outside the parabola where the point (-2, 1) lies. 𝑦 < 𝑥2 2 3 − 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution. Therefore, all the points in the shaded region are the solutions of 𝑦 < 𝑥2 − 2𝑥 − 3. (-2,0) (4,5) (6,3) To many to mention
• 40.
  DAY 03
• 41.
  SHORT REVIEW
• 42.
  Is the point(3,0) a solution of the inequality y x ≥ 2 5x + 2? − NO YES
• 43.
  Is the point(0,5) a solution of the inequality y < -x2 + 5? YES NO
• 44.
  Is the point(0,0) a solution of the inequality y (x - 5)(x + 3)? ≤ YES NO
• 45.
  WORKED EXAMPLE
• 46.
  STEP 1: Rewritethe inequality as a quadratic equation where the right side is 0. 𝑥2 + 6 7= 0 𝑥 − STEP 2: Solve the equation by factoring, quadratic formula, or completing the square as the case may be and find its roots. 𝑥2 + 6 7 = 0 𝑥 − ( + 7)( - 1) = 0 𝑥 𝑥 𝑥 = -7, = 1 𝑥 The roots of the equation are (-7, 0) and (1, 0). 𝑦 ≥ 𝑥2 + 6 7 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution.
• 47.
  STEP 3: Findthe vertex of the equation. 𝑦 ≥ 𝑥2 + 6 7 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution. Vertex = (h, k) = = = = = The vertex is (-3, -16).
• 48.
  The roots ofthe equation are (-7, 0) and (1, 0) The vertex is (-3, -16). 𝑦 ≥ 𝑥2 + 6 7 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution.
• 49.
  STEP 4: Selecta point ( , ) in each 𝑥 𝑦 region (inside and outside the parabola) and check whether the given inequality is satisfied 𝑦 ≥ 𝑥2 + 6 7 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution. Test point (0, 0) which is found inside the parabola. 𝑦 ≥ 𝑥2 + 6 7 𝑥 − 0 0 ≥ 2 + 6(0) 7 − 0 0 + 0 7 ≥ − 0 7 ≥ − TRUE Test point (2, -2) which is found outside the parabola. 𝑦 ≥ 𝑥2 + 6 7 𝑥 − -2 2 ≥ 2 + 6(2) 7 − -2 4 + 12 7 ≥ − -2 9 ≥ FALSE
• 50.
  STEP 5: Shadethe region that contains the solution. Shade the region inside the parabola where the point (0, 0) lies. 𝑦 < 𝑥2 2 3 − 𝑥 − Graph the following quadratic inequality and shade the region that contains the solution. Therefore, all the points in the shaded region are the solutions of ≥ 𝑦 𝑥2 + 6 + 𝑥 7.
• 51.
  substitute 1 and4 to and , respectively. 𝑥 𝑦 𝑦 ≤ 6𝑥2 13 28 − 𝑥 − 4 6(1) ≤ 2 13(1) 28 − − 4 6(1) 13 28 ≤ − − 4 6 13 28 ≤ − − 4 35 ≤ − FALSE Since the given point doesn’t satisfy the inequality, 𝐴(1,4) is not a solution 𝑦 ≤ 6𝑥2 13 28 − 𝑥 − Determine whether (1,4) is a solution of 𝐴
• 52.
  Substitute 2 and-5 to x and y, respectively. 𝑦 > 6𝑥2 11 30 − 𝑥 − −5 > 6(2)2 11(2) 30 − − −5 > 6(4) 22 30 − − −5 > 24 22 30 − − −5 > 28 − Since the given point satisfies the inequality, (2, 5) is a 𝐵 − solution. 𝑦 > 6𝑥2 11 30. − 𝑥 − Determine whether (2, 5) is a solution of 𝐵 −
• 53.
  DAY 04
• 54.
  LESSON ACTIVITY
• 55.
  Activity 1: DoI Belong Here? Write YES if the point is a solution of the inequality. Otherwise, write NO. In addition, ask them to provide a short justification of their answers. The items are the following: 1) > 𝑦 𝑥2 7 9 − − 𝑥 (4,-2) 2) < ( + 6) + 5 𝑦 𝑥 𝑥 (-6, ½) 3) 5 𝑦 ≥ 𝑥2 17 + 6 − 𝑥 (2, -6) 4) 5 > 2 𝑦 − 𝑥2 + 11𝑥 (-4, 5 /2) 5) (3 10) + 3 𝑦 ≤ 𝑥 𝑥− (1, 8)
• 56.
  Activity 2: SketchMy Graph 1) ≥ 𝑥 ≥ 2 6𝑥 + 5 − 2) 𝑦 < 2𝑥2 3𝑥 + 1 −
• 57.
  Activity 1: DoI Belong Here? Write YES if the point is a solution of the inequality. Otherwise, write NO. In addition, ask them to provide a short justification of their answers. The items are the following: 1) > 𝑦 𝑥2 7 9 − − 𝑥 (4,-2) 2) < ( + 6) + 5 𝑦 𝑥 𝑥 (-6, ½) 3) 5 𝑦 ≥ 𝑥2 17 + 6 − 𝑥 (2, -6) 4) 5 > 2 𝑦 − 𝑥2 + 11𝑥 (-4, 5 /2) 5) (3 10) + 3 𝑦 ≤ 𝑥 𝑥− (1, 8) YES YES YES YES NO ANSWER KEY
• 58.
  Activity 2: SketchMy Graph 1) 𝑦 ≥ 𝑥2 6𝑥 + 5 − 2) < 2 𝑦 𝑥2 3𝑥 + − 1 ANSWER KEY