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Comprehensive Guide to Quadratic Inequalities in Two Variables with Graphing Techniques

Explore quadratic inequalities in two variables, their forms, graphing methods, vertex calculation, and solution shading on the Cartesian plane with practical examples and activities.

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Math 10, Term 1, Week 6
A quadratic inequality in 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
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
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
A dash or broken 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.
The Parabola opens upward if 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
VERTEX The turning point. Also, the highest or lowest point. If the given is an inequality, use the formula Vertex = (h,k) =
STEP 1: Rewrite the 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.
STEP 3: Find the 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).
The roots of the 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.
STEP 4: Select a 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
STEP 5: Shade the 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
STEP 1: Rewrite the 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.
STEP 3: Find the 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).
The roots of the 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.
STEP 4: Select a 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
STEP 5: Shade the region that contains the solution. Shade the region inside the parabola where the point (0, 0) lies. 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. 𝑦 ≥ 𝑥2 + 6 7 𝑥 −
substitute 1 and 4 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
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 −
DAY 04
LESSON ACTIVITY
Activity 1: Do I Belong Here? By Pair. Activity notebook. 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)
Activity 2: Sketch My Graph. By Pair. Activity notebook. Copy and answer with solution 1) 𝑦 ≥ 𝑥2 6 + 5 − 𝑥 2) 𝑦 < 2𝑥2 3 + 1 − 𝑥