Algebra 2 Questions Urgent

Question

algebraically write the equation of a parabola in standard form (y=ax^2+bx+c) that passses through the points (-1,6)(1,2)(2,9)

Answer 1

Take the 3 points and plug them into the general equation, and this will give you 3 equations for a, b and c (3 equations, 3 unknowns):

(-1,6):   6 = a * (-1)^2 + b * (-1) + C   this becomes: 6=a-b+c     equation 1

(1,2): 2 = a * (1^2) + b*(1) + c:         this becomes: 2=a+b+c     equation 2

(2,9): 9=a*(2^2) + b*(2) + c:             this becomes: 9=4a+2b+c  equation 3

 

calculate equation 2 minus equation 1:  2-6 = 2*b, or b=-2

Substitute this back into equations 1 and 3: 

6=a+2+c   this becomes: 4=a+c

9=4a-4+c  thi becomes: 13 = 4a+c

 

These two equations can then be solved to give a=3, c=1

 

so the equation is: y=3x^2-2x+1