How do you write y=3x^2+x-6 in the form of y=a(x-h)^2+k

Question

I’ve tried myself and have looked at dif sites to help but can’t find the answer

Answer 1

The question is to rewrite the standard from into vertex form. This can be done using Completing the Square.

 

3x²+x -6 =0

3[x² + (1/3)x -2 ]=0                               ...factor 3 out

3[x²+ (1/3)x + (1/36) - 2 - (1/36)]=0  ... complete the square  [b/(2a)]² = (1/6)²

3[(x + 1/6)² - 73/36]= 0                       ...simplify

3(6x + 1)²- 73/12= 0                              ...distribute 3