1 / 24

Writing Quadratic Functions in Different Forms

Learn how to write quadratic functions in vertex form, intercept form, and standard form with step-by-step examples and practice problems.

dmckinney
Download Presentation

Writing Quadratic Functions in Different Forms

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Write a quadratic function in vertex form EXAMPLE 1 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 forhand –2 for k. Use the other given point, (3, 2), to find a. 2= a(3– 1)2 – 2 Substitute 3 for xand 2 for y. 2 = 4a – 2 Simplify coefficient of a. 1 = a Solve for a.

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

  3. Write a quadratic function in intercept form EXAMPLE 2 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 pand 4 for q.

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

  5. Write a quadratic function in standard form EXAMPLE 3 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 = ax2 +bx + cto obtain the system of three linear equations shown below.

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

  7. Write a quadratic function in standard form EXAMPLE 3 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. 6a = 12 ANSWER A quadratic function for the parabola is y = 2x2 + x – 4. Write a quadratic function in standard form EXAMPLE 3 2a – 2b = 2 a – b = 1 4a + 2b = 10 4a + 2b = 10 a = 2 So 2 – b = 1, which means b = 1. The solution is a = 2, b = 1, and c = – 4.

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

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

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

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

  13. 3.x-intercepts: –2, 5 passes through: (6, 2) for Examples 1, 2 and 3 GUIDED PRACTICE 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 pand –5 for q.

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

  15. for Examples 1, 2 and 3 GUIDED PRACTICE 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 = ax2 +bx + cto obtain the system of three linear equations shown below.

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

  17. for Examples 1, 2 and 3 GUIDED PRACTICE 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. 6a = 24 ANSWER A quadratic function for the parabola is 4x2– 2x – 1 y = for Examples 1, 2 and 3 GUIDED PRACTICE a – b = 6 2a – 2b = 12 4a + 2b = 10 4a + 2b = 12 a = 4 So, 4 – b = 6, which means b = – 2

  19. for Examples 1, 2 and 3 GUIDED PRACTICE 5. (–2, –1), (0, 3), (4, 1) SOLUTION STEP 1 Substitute the coordinates of each point into y = ax2 +bx + c = 0to obtain the system of three linear equations shown below.

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

  21. for Examples 1, 2 and 3 GUIDED PRACTICE –1 = 4a – 2b + c Equation 1 4a – 2b + 3 = – 1 Substitute 3 for c. 4a – 2b = – 1 Revised Equation 1 1 = 16a + 4b + c Equation 3 16a + 4b + 3= 1 Substitute 3 for c. 16a2+ 4b = –2 Revised Equation 3 STEP 3 Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

  22. 24a = –10 5 5 5 a = 7 7 12 12 12 6 6 So 16 +4b = , which means b = . ANSWER – 5 A quadratic function for the parabola is y = x2 + x + 3. 12 for Examples 1, 2 and 3 GUIDED PRACTICE 4a – 2b = – 4 8a – 4b = – 8 16a + 4b = – 2 16a + 4b = – 2

  23. for Examples 1, 2 and 3 GUIDED PRACTICE 6. (–1, 0), (1, –2), (2, –15) SOLUTION STEP 1 Substitute the coordinates of each point into y = ax2 +bx + c to obtain the system of three linear equations shown below.

  24. for Examples 1, 2 and 3 GUIDED PRACTICE ANSWER A quadratic function for the parabola is y = 4x2x + 3

More Related