CONVERTING FROM GENERAL FORM TO STANDARD FORM FOR PARABOLA

Write the following equations in standard form :

Problem 1 :

y = 2x² - 12 x + 6

Solution :

y = 2x² - 12 x + 6

Factoring 2 from x terms,

y = 2(x² - 6 x) + 6

y = 2[x² - 2 x (3) + 3² - 3²] + 6

y = 2[(x - 3)² - 9] + 6

Distributing 2, we get

y = 2(x - 3)² - 18 + 6

y = 2(x - 3)² - 12

Problem 2 :

y = x² - 6 x + 11

Solution :

y = x² - 6 x + 11

y = [x² - 2 x (3) + 3² - 3²] + 11

y = [(x - 3)² - 3²] + 11

y = [(x - 3)² - 9] + 11

y = (x - 3)² - 9 + 11

y = (x - 3)² + 2

Problem 3 :

x = y² + 14y + 20

Solution :

x = y² + 14y + 20

x = [y² + 2 x (7) + 7² - 7²] + 20

x = [(y + 7)² - 7²] + 20

x = (y + 7)² - 49 + 20

x = (y + 7)² - 29

Problem 4 :

y = (1/2) x² + 12x - 8

Solution :

y = (1/2) x² + 12x - 8

Factoring 1/2 from x² and x, we get

y = (1/2)[x² + 24x] - 8

y = (1/2)[x² + 2 x (12) + 12² - 12²] - 8

y = (1/2)[(x + 12)²  - 144] - 8

Distributing 1/2, we get

y = (1/2)(x + 12)²  - 72 - 8

y = (1/2)(x + 12)²  - 80

Problem 5 :

x = 3y² + 5y - 9

Solution :

x = 3y² + 5y - 9

Factoring 3 from y terms it is not possible, because the coefficient of y is not a multiple of 3.

x = 3[y² + (5/3)y] - 9

Writing the coefficient of x as multiple of 2, we get

x = 3[y² + 2(5/6)y + (5/6)² - (5/6)²] - 9

x = 3[(y + (5/6))² - (25/36)] - 9

x = 3(y + (5/6))² - (25/12) - 9

x = 3(y + (5/6))² + (-25 - 108)/12

x = 3(y + (5/6))² + (-133)/12

Problem 6 :

Katie is finding the standard form of the equation

y = x² + 6x + 4

Solution :

y = x² + 6x + 4

y = x² + 2 x (3) + 3² - 3² + 4

y = (x + 3)² - 3² + 4

y = (x + 3)² - 9 + 4

y = (x + 3)² - 5