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1. 1. 6.7 Graph and Solving Quadratic Inequalities
2. 2. Method of Graph sketching
3. 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 <ul><li>Graphs will look like a parabola with a solid or dotted line and a shaded section. </li></ul><ul><li>The graph could be shaded inside the parabola or outside. </li></ul>
4. 4. Steps for graphing <ul><li>1. Sketch the parabola y=ax 2 +bx+c </li></ul><ul><li>(dotted line for < or >, solid line for ≤ or ≥) </li></ul><ul><li>** remember to use 5 points for the graph! </li></ul><ul><li>2. Choose a test point and see whether it is a solution of the inequality. </li></ul><ul><li>3. Shade the appropriate region. </li></ul><ul><li>(if the point is a solution, shade where the point is, if it’s not a solution, shade the other region) </li></ul>
5. 5. Example: Graph y ≤ x 2 +6x- 4 * Vertex: (-3,-13) * Opens up, solid line <ul><li>Test Point: (0,0) </li></ul><ul><li>0 ≤0 2 +6(0)-4 </li></ul><ul><li>0≤-4 </li></ul>So, shade where the point is NOT! Test point
6. 6. Graph: y>-x 2 +4x-3 <ul><li>* Opens down, dotted line. </li></ul><ul><li>* Vertex: (2,1) </li></ul>* Test point (0,0) 0>-0 2 +4(0)-3 0>-3 <ul><li>x y </li></ul><ul><li>0 -3 </li></ul><ul><li>1 0 </li></ul><ul><li>1 </li></ul><ul><li>0 </li></ul><ul><li>-3 </li></ul>Test Point
7. 7. Last Example! Sketch the intersection of the given inequalities. 1 y ≥x 2 and 2 y≤-x 2 +2x+4 <ul><li>Graph both on the same coordinate plane. The place where the shadings overlap is the solution. </li></ul><ul><li>Vertex of #1: (0,0) </li></ul><ul><li>Other points: (-2,4), (-1,1), (1,1), (2,4) </li></ul><ul><li>Vertex of #2: (1,5) </li></ul><ul><li>Other points: (-1,1), (0,4), (2,4), (3,1) </li></ul><ul><li>* Test point (1,0): doesn’t work in #1, works in #2. </li></ul>SOLUTION!
9. 9. Solve the quadratic inequality x 2 – 5 x + 6 > 0 graphically. Example 1:
10. 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. 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. 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. 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. 14. From the sketch, we obtain the solution or
15. 15. Graphical Solution: 0 2 3
16. 16. Solve the quadratic inequality x 2 – 5 x + 6 < 0 graphically. Example 2: Same method as example 1 !!!
17. 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. 18. From the sketch, we obtain the solution 2 < x < 3
19. 19. Graphical Solution: 0 2 3
20. 20. <ul><li>Solve </li></ul>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. 21. <ul><li>Solve </li></ul>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. 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. 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