How can I find the equation in standard form of a parabola that passes through the points (-1,-12) (0,-6) (3,0)

Question

I'm lost on this question, can someone save me?;-;

Answer 1

y = Ax^2 + Bx + C

 

 

(3,0) --->   A(9) + B(3) + C = 0

 

                 9A + 3B + C = 0

 

(0,-6) --->   C = -6

 

substituting into the first equation:

 

  9A + 3B - 6 = 0

 

  3A + B - 2 = 0

 

  3A + B = 2

 

   B = 2 - 3A

 

(-1,-12) --->    A(-1)^2 + B(-1) + -6 = -12

 

                       A - B + -6= -12

 

                        A - B = -6

 

A - B = -6

 

A - (2 - 3A) = -6

 

A - 2 + 3A = -6

 

4A - 2 = -6

 

4A = -4

 

A = -1

 

B  = 2 - 3A = 2 - 3(-1) = 2 + 3 = 5

 

-x^2 + 5x - 6

 

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