1. 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.

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

3. 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.

4. 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

5. 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.

6. 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

7. 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.

8. 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

9. GUIDED PRACTICE for Examples 1, 2 and 3 Write a quadratic function whose graph has the given characteristics. 1. vertex: (4, –5) passes through: (2, –1)

10. GUIDED PRACTICE for Examples 1, 2 and 3 SOLUTION Use vertex form because the vertex is given. y = a(x – h) 2 + k Vertex form y = a(x – 4) 2 – 5 Substitute 4 for h and –5 for k. Use the other given point, (2,–1), to find a. –1 = a(2 – 4) 2 – 5 Substitute 2 for x and –1 for y. –1 = 4a – 5 Simplify coefficient of x. –1 = a Solve for a. A quadratic function for the parabola is y = (x – 4) 2 – 5. ANSWER

11. GUIDED PRACTICE for Examples 1, 2 and 3 2. vertex: (–3, 1) passes through: (0, –8)

12. GUIDED PRACTICE for Examples 1, 2 and 3 SOLUTION Use vertex form because the vertex is given. y = a(x – h) 2 + k Vertex form y = a(x + 3) 2 + 1 Substitute –3 for h and 1 for k. Use the other given point, (2,–1), to find a. –8 = a(0 + 3) 2 + 1 Substitute 2 for x and –8 for y. –8 = 9a + 1 Simplify coefficient of x. –1 = a Solve for a. A quadratic function for the parabola is y =  (x + 3) 2 + 1. ANSWER

13. GUIDED PRACTICE for Examples 1, 2 and 3 3. x -intercepts: –2, 5 passes through: (6, 2) SOLUTION Use intercept form because the x -intercepts are given. y = a(x – p)(x – q) Intercept form y = a(x + (– 2))(x – 5) Substitute –2 for p and –5 for q.

14. GUIDED PRACTICE for Examples 1, 2 and 3 Use the other given point, (6, 2), to find a. 2 = a(6 + (–2))(6 – 5) Substitute 6 for a and 2 for y. 2 = 8a Simplify coefficient of a. Solve for a. 1 4 = a A quadratic function for the parabola is 1 4 (x + 2)(x – 5).y = ANSWER

15. GUIDED PRACTICE for Examples 1, 2 and 3 Write a quadratic function in standard form for the parabola that passes through the given points. 4. (–1, 5), (0, –1), (2, 11) 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.

16. GUIDED PRACTICE for Examples 1, 2 and 3 5 = a(–1) 2 + b(–1) + c Substitute –1 for x and 5 for y. 5 = a – b + c Equation 1 –1 = a(0) 2 + b(0) + c Substitute 0 for x and – 1 for y. – 1 = c Equation 2 11 = a(2) 2 + b(2) + c Substitute 2 for x and 11 for y. 11 = 4a + 2b + c Equation 3 Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 1 for c in Equations 1 and 3. STEP 2

17. GUIDED PRACTICE for Examples 1, 2 and 3 a – b + c = 5 Equation 1 a – b – 1 = 5 Substitute – 1 for c. a – b = 6 Revised Equation 1 4a + 2b + c = 11 Equation 3 4a + 2b – 1 = 11 Substitute – 1 for c. 4a + 2b = 12 Revised Equation 3 STEP 3 Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

18. GUIDED PRACTICE for Examples 1, 2 and 3 a – b = 62a – 2b = 12 4a + 2b = 104a + 2b = 12 6a = 24 a = 4 So, 4 – b = 6, which means b = – 2 A quadratic function for the parabola is 4x 2 – 2x – 1 y = ANSWER

19. GUIDED PRACTICE for Examples 1, 2 and 3 5. (–2, –1), (0, 3), (4, 1) SOLUTION STEP 1 Substitute the coordinates of each point into y = ax 2 + bx + c = 0 to obtain the system of three linear equations shown below.

20. GUIDED PRACTICE for Examples 1, 2 and 3 Rewrite the system of three equations in Step 1 as a system of two equations by substituting 3 for c in Equations 1 and 3. STEP 2 –1 = a(–2) 2 + b(–2) + c Substitute –2 for y and –1 for x. –1 = 4a – 2b + c Equation 1 3 = a(0) 2 + b(0) + c Substitute 0 for x and 3 for y. 3 = c Equation 2 1 = a(4) 2 + b(4) + c Substitute 1 for y and 4 for x. 1 = 16a + 4b + c Equation 3

21. GUIDED PRACTICE for Examples 1, 2 and 3 Equation 1 4a – 2b + 3 = – 1 Substitute 3 for c. Revised Equation 1 Equation 3 16a + 4b + 3 = 1 Substitute 3 for c. 16a 2 + 4b = –2 Revised Equation 3 STEP 3 Solve the system consisting of revised Equations 1 and 3. Use the elimination method. –1 = 4a – 2b + c 4a – 2b = – 1 1 = 16a + 4b + c

22. GUIDED PRACTICE for Examples 1, 2 and 3 4a – 2b = – 48a – 4b = – 8 16a + 4b = – 2 24a = –10 a = 5 12 So 16 +4b =, which means b =. 5 12 5 7 6 ANSWER A quadratic function for the parabola is y = x 2 + x + 3. – 5 12 7 6

23. GUIDED PRACTICE for Examples 1, 2 and 3 6. (–1, 0), (1, –2), (2, –15) 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.

24. GUIDED PRACTICE for Examples 1, 2 and 3 ANSWER A quadratic function for the parabola is y =  4x 2  x + 3