1. Chapter 1 : Functions Sept 29, 2005

2. Solving Quadratic Inequalities Graphically 1.Write in standard form F(x) > 0. 2.Factor if possible. Find zeros of F(x). 3. Determine if parabola points up or down. up if a is positive, down if a is negative. 4. Sketch F(x). 5. Solve the inequality by looking at the graph and determining the appropriate region:

3. Note: When solving for; F(x) < 0 : Find x-values where graph is below x-axis F(x) > 0: Find x-values where graph is above x-axis F(x) < 0 : Find x-values where graph is below or on x-axis F(x) > 0: Find x-values where graph is above or on x-axis

4. Solving a Quadratic Inequality Example 1: Solve 2x 2 -2x -4 > 0. Find the zero’s of F(x) = 2x 2 -2x -4 Step 3: Sketch f(x) = 2x 2 -2x -4 The interval for f(x) > 0 is when the graph is above the x-axis. Therefore, solution is ] -∞, -1[ U ]2, ∞[ Step2: Step1: Work: 2x 2 -2x -4=0 2(x 2 -x -2)=0 (x-2) (x+1)=0 x =2, x= -1 Graph points upward : a is +ve The solution is where the graph of F(x) is above the x-axis

5. Solving a Quadratic Inequality Solve x 2 + x < 20. Write in standard form: x 2 + x - 20 < 0 The solution is ]-5, 4 [ x > -5 and x < 4 The interval for f(x) < 0 is when the graph is below the x-axis. Step1: Find the zero’s of F(x) = x 2 +x -20 Work: x 2 +x -20=0 (x+5)(x-4)=0 x =-5, x= 4 Step2:Graph points up since a is +ve Step 3: Sketch f(x) = x 2 +x -20

6. Solve x 2 + x > 20. Graph as f(x) = x 2 + x - 20 The interval for f(x) > 0 is when the graph is above the x-axis. The solution is ]- ,-5[ U ]4,  [ x<-5 x > 4 Solving a Quadratic Inequality Write in standard form: x 2 + x - 20 > 0

7. Solving a Quadratic Inequality Solve x 2 –8x + 16 ≥ 9. Write in standard form: x 2 –8x + 7 ≥ 0 Graph as f(x) = x 2 – 8x+7 The zero’s of x 2 – 8x+7 are: ( x – 7) (x –1)=0, x= 7 x= 1 The interval for f(x) ≥ 0 is when the graph is above or on the x-axis. The solution is ]- ,1] U [7,  [

8. Solving a Quadratic Inequality Solve x 2 +6x+10 ≥-x 2 -6x-8 Write in standard form: 2x 2 +12x + 18 ≥ 0 Graph f(x) = 2x 2 +12x+18 The zero’s of 2x 2 +12x + 18 are: 2(x 2 +6x+9)=0 x 2 +6x+9=0 (x+3)(x+3)=0 x=-3 The interval for f(x) ≥ 0 is when the graph is above or on the x-axis. The solution is ]- ,  [ which means everywhere.

9. Page 59 #10(b,c), 20, 28(a)