6.7 quadratic inequalities
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This document discusses how to graph and solve quadratic inequalities. It provides steps for graphing quadratic inequalities by sketching the parabola and shading the appropriate region based on a test point. Examples are given of solving quadratic inequalities graphically by determining the portions of the graph above or below the x-axis and obtaining the solution intervals. Exercises are also worked through to practice solving quadratic inequalities graphically.
6.7 quadratic inequalities
- 1. 6.7 Graph and Solving Quadratic Inequalities
- 2. Method of Graph sketching
- 3. Forms of Quadratic Inequalities y<ax 2 +bx+c y>ax 2 +bx+c y ≤ax 2 +bx+c y≥ax 2 +bx+c Graphs will look like a parabola with a solid or dotted line and a shaded section. The graph could be shaded inside the parabola or outside.
- 4. Steps for graphing 1. Sketch the parabola y=ax 2 +bx+c (dotted line for < or >, solid line for ≤ or ≥) ** remember to use 5 points for the graph! 2. Choose a test point and see whether it is a solution of the inequality. 3. Shade the appropriate region. (if the point is a solution, shade where the point is, if it’s not a solution, shade the other region)
- 5. Example: Graph y ≤ x 2 +6x- 4 * Vertex: (-3,-13) * Opens up, solid line Test Point: (0,0) 0 ≤0 2 +6(0)-4 0≤-4 So, shade where the point is NOT! Test point
- 6. Graph: y>-x 2 +4x-3 * Opens down, dotted line. * Vertex: (2,1) * Test point (0,0) 0>-0 2 +4(0)-3 0>-3 x y 0 -3 1 0 1 0 -3 Test Point
- 7. Last Example! Sketch the intersection of the given inequalities. 1 y ≥x 2 and 2 y≤-x 2 +2x+4 Graph both on the same coordinate plane. The place where the shadings overlap is the solution. Vertex of #1: (0,0) Other points: (-2,4), (-1,1), (1,1), (2,4) Vertex of #2: (1,5) Other points: (-1,1), (0,4), (2,4), (3,1) * Test point (1,0): doesn’t work in #1, works in #2. SOLUTION!
- 9. Solve the quadratic inequality x 2 – 5 x + 6 > 0 graphically. Example 1:
- 10. Procedures: Step (2): we have y = ( x – 2)( x – 3) , i.e. y = 0, when x = 2 or x = 3. Factorize x 2 – 5 x + 6, The corresponding quadratic function is y = x 2 – 5 x + 6 Sketch the graph of y = x 2 – 5 x + 6. Step (1): Step (3): Step (4): Find the solution from the graph.
- 11. Sketch the graph y = x 2 – 5 x + 6 . What is the solution of x 2 – 5 x + 6 > 0 ? x y 0 y = ( x – 2)( x – 3) , y = 0, when x = 2 or x = 3. 2 3
- 12. x y 0 We need to solve x 2 – 5 x + 6 > 0, The portion of the graph above the x-axis represents y > 0 (i.e. x 2 – 5 x + 6 > 0) The portion of the graph below the x-axis represents y < 0 (i.e. x 2 – 5 x + 6 < 0) above the x-axis. so we choose the portion 2 3
- 13. x y 0 When x < 2 , the curve is above the x-axis i.e., y > 0 x 2 – 5x + 6 > 0 When x > 3 , the curve is above the x-axis i.e., y > 0 x 2 – 5x + 6 > 0 2 3
- 14. From the sketch, we obtain the solution or
- 15. Graphical Solution: 0 2 3
- 16. Solve the quadratic inequality x 2 – 5 x + 6 < 0 graphically. Example 2: Same method as example 1 !!!
- 17. x y 0 When 2 < x < 3 , the curve is below the x-axis i.e., y < 0 x 2 – 5 x + 6 < 0 2 3
- 18. From the sketch, we obtain the solution 2 < x < 3
- 19. Graphical Solution: 0 2 3
- 20. Solve Exercise 1 : x < –2 or x > 1 Answer: Find the x-intercepts of the curve: (x + 2)(x – 1)=0 x = –2 or x = 1 x y 0 0 – 2 1 – 2 1
- 21. Solve Exercise 2 : – 3 < x < 4 Answer: Find the x-intercepts of the curve: x 2 – x – 12 = 0 (x + 3)(x – 4)=0 x = –3 or x = 4 x y 0 0 – 3 4 – 3 4
- 22. Solve Exercise 3 : – 7 < x < 5 Solution: Find the x-intercepts of the curve: (x + 7)(x – 5)=0 x = –7 or x = 5 x y 0 0 – 7 5 – 7 5
- 23. Solve Exercise 4 : Solution: Find the x-intercepts of the curve: (x + 3)(3x – 2)=0 x = –3 or x = 2/3 x –3 or x 2/3 x y 0 – 3 2 3 0 – 3 2 3