1. EXAMPLE 1 Solve an equation with two real solutions Solve x 2 + 3x = 2. x 2 + 3x = 2 Write original equation. x 2 + 3x – 2 = 0 Write in standard form. x = – b + b 2 – 4ac 2a2a Quadratic formula x = – 3 + 3 2 – 4(1)( –2) 2(1) a = 1, b = 3, c = –2 Simplify. x = – 3 +– 3 + 17 2 The solutions are x = – 3 +– 3 + 17 2 0.56 and x = – 3 –– 3 – 17 2 – 3.56. ANSWER

2. EXAMPLE 1 Solve an equation with two real solutions CHECK Graph y = x 2 + 3x – 2 and note that the x -intercepts are about 0.56 and about – 3.56. 

3. EXAMPLE 2 Solve an equation with one real solutions Solve 25x 2 – 18x = 12x – 9. 25x 2 – 18x = 12x – 9. Write original equation. Write in standard form. x = 30 + (–30) 2 – 4(25)(9) 2(25) a = 25, b = –30, c = 9 Simplify. 25x 2 – 30x + 9 = 0. x = 30 + 0 50 x = 3 5 Simplify. 3 5 The solution is ANSWER

4. EXAMPLE 2 Solve an equation with one real solutions CHECK Graph y = –5x 2 – 30x + 9 and note that the only x -intercept is 0.6 =. 3 5 

5. EXAMPLE 3 Solve an equation with imaginary solutions Solve –x 2 + 4x = 5. –x 2 + 4x = 5 Write original equation. Write in standard form. x = – 4+  4 2 – 4(– 1)(– 5) 2(– 1) a = –1, b = 4, c = –5 Simplify. –x 2 + 4x – 5 = 0. x = – 4+ – 4 – 2 – 4+ 2i x = – 2 Simplify. Rewrite using the imaginary unit i. x = 2 + i The solution is 2 + i and 2 – i. ANSWER

6. EXAMPLE 3 Solve an equation with imaginary solutions CHECK Graph y = 2x 2 + 4x – 5. There are no x - intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + i is shown. –(2 + i) 2 + 4(2 + i) = 5 ? –3 – 4i + 8 + 4i = 5 ? 5 = 5 

7. EXAMPLE 4 Use the discriminant Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. a. x 2 – 8x + 17 = 0 b. x 2 – 8x + 16 = 0 c. x 2 – 8x + 15 = 0 SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac x = – b+ b 2 – 4ac 2ac a. x 2 – 8x + 17 = 0(–8) 2 – 4(1)(17) = – 4 Two imaginary: 4 + i b. x 2 – 8x + 16 = 0(–8) 2 – 4(1)(16) = 0 One real: 4 b. x 2 – 8x + 15 = 0(–8) 2 – 4(1)(15) = 0 Two real: 3,5

8. EXAMPLE 1 Graph a quadratic inequality Graph y > x 2 + 3x – 4. SOLUTION STEP 1 Graph y = x 2 + 3x – 4. Because the inequality symbol is >, make the parabola dashed. Test a point inside the parabola, such as (0, 0). STEP 2 y > x 2 + 3x – 4 0 > 0 2 + 3(0) – 4 ? 0 > – 4

9. EXAMPLE 1 Graph a quadratic inequality So, (0, 0) is a solution of the inequality. STEP 3 Shade the region inside the parabola.

10. EXAMPLE 2 Use a quadratic inequality in real life A manila rope used for rappelling down a cliff can safely support a weight W (in pounds) provided Rappelling W ≤ 1480d 2 where d is the rope’s diameter (in inches). Graph the inequality. SOLUTION Graph W = 1480d 2 for nonnegative values of d. Because the inequality symbol is ≤, make the parabola solid. Test a point inside the parabola, such as (1, 2000).

11. EXAMPLE 2 Use a quadratic inequality in real life W ≤ 1480d 2 2000 ≤ 1480(1) 2 2000 ≤ 1480 Because (1, 2000) is not a solution, shade the region below the parabola.

12. EXAMPLE 3 Graph a system of quadratic inequalities Graph the system of quadratic inequalities. y < – x 2 + 4 Inequality 1 y > x 2 – 2x – 3 Inequality 2 SOLUTION STEP 1 Graph y ≤ – x 2 + 4. The graph is the red region inside and including the parabola y = – x 2 + 4.

13. EXAMPLE 3 Graph a system of quadratic inequalities STEP 2 Graph y > x 2 – 2x – 3. The graph is the blue region inside (but not including) the parabola y = x 2 –2x – 3. Identify the purple region where the two graphs overlap. This region is the graph of the system. STEP 3

14. EXAMPLE 4 Solve a quadratic inequality using a table Solve x 2 + x ≤ 6 using a table. SOLUTION Rewrite the inequality as x 2 + x – 6 ≤ 0. Then make a table of values. Notice that x 2 + x –6 ≤ 0 when the values of x are between –3 and 2, inclusive. The solution of the inequality is –3 ≤ x ≤ 2. ANSWER

15. EXAMPLE 5 Solve a quadratic inequality by graphing Solve 2x 2 + x – 4 ≥ 0 by graphing. SOLUTION The solution consists of the x -values for which the graph of y = 2x 2 + x – 4 lies on or above the x -axis. Find the graph’s x -intercepts by letting y = 0 and using the quadratic formula to solve for x. 0 = 2x 2 + x – 4 x = – 1+ 1 2 – 4(2)(– 4) 2(2) x = – 1+ 33 4 x 1.19 or x –1.69

16. EXAMPLE 5 Solve a quadratic inequality by graphing Sketch a parabola that opens up and has 1.19 and –1.69 as x -intercepts. The graph lies on or above the x -axis to the left of (and including) x = – 1.69 and to the right of (and including) x = 1.19. The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19. ANSWER

17. EXAMPLE 1 Write a quadratic function in vertex form Write a quadratic function for the parabola shown. SOLUTION Use vertex form because the vertex is given. y = a(x – h) 2 + k Vertex form y = a(x – 1) 2 – 2 Substitute 1 for h and –2 for k. Use the other given point, (3, 2), to find a. 2 = a(3 – 1) 2 – 2 Substitute 3 for x and 2 for y. 2 = 4a – 2 Simplify coefficient of a. 1 = a Solve for a.

18. EXAMPLE 1 Write a quadratic function in vertex form A quadratic function for the parabola is y = (x – 1) 2 – 2. ANSWER

19. EXAMPLE 2 Write a quadratic function in intercept form Write a quadratic function for the parabola shown. SOLUTION Use intercept form because the x -intercepts are given. y = a(x – p)(x – q) Intercept form y = a(x + 1)(x – 4) Substitute –1 for p and 4 for q.

20. EXAMPLE 2 Write a quadratic function in intercept form Use the other given point, (3, 2), to find a. 2 = a(3 + 1)(3 – 4) Substitute 3 for x and 2 for y. 2 = – 4a Simplify coefficient of a. Solve for a. 1 2 – = a A quadratic function for the parabola is 1 2 – (x + 1)(x – 4).y = ANSWER

21. EXAMPLE 3 Write a quadratic function in standard form Write a quadratic function in standard form for the parabola that passes through the points (–1, –3), (0, – 4), and (2, 6). SOLUTION STEP 1 Substitute the coordinates of each point into y = ax 2 + bx + c to obtain the system of three linear equations shown below.

22. EXAMPLE 3 Write a quadratic function in standard form –3 = a(–1) 2 + b(–1) + c Substitute –1 for x and 23 for y. –3 = a – b + c Equation 1 –3 = a(0) 2 + b(0) + c Substitute 0 for x and – 4 for y. – 4 = c Equation 2 6 = a(2) 2 + b(2) + c Substitute 2 for x and 6 for y. 6 = 4a + 2b + c Equation 3 Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 4 for c in Equations 1 and 3. STEP 2

23. EXAMPLE 3 Write a quadratic function in standard form a – b + c = – 3 Equation 1 a – b – 4 = – 3 Substitute – 4 for c. a – b = 1 Revised Equation 1 4a + 2b + c = 6 Equation 3 4a + 2b + 4 = 6 Substitute – 4 for c. 4a + 2b = 10 Revised Equation 3 STEP 3 Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

24. EXAMPLE 3 Write a quadratic function in standard form a – b = 1 2a – 2b = 2 4a + 2b = 10 6a = 12 a = 2 So 2 – b = 1, which means b = 1. The solution is a = 2, b = 1, and c = – 4. A quadratic function for the parabola is y = 2x 2 + x – 4. ANSWER