Axis of symmetry: x=-3; points: (-7,-2), (-5,-8); direction of opening: upward

Answer 1

y = Ax^2 + bx + C <-- 3 Undetermined Coefficients, so 3 equations are required

Axis of symmetry is x=-3

So then -b/(2a) = -3 --> -b = -6a --> b = 6a is the first equation

(-7,-2)

-2 = 49a -7b + c is the 2nd equation

(-5,-8)

-8 = 25a - 5b + c is the 3rd equation

3 by 3 system is

b = 6a

-2 = 49a -7b + c

-8 = 25a - 5b + c

Eliminates c by subtracting the second equation MINUS the 3rd equation:

-2 - (-8) = 49a - 25a - 7b - (-5b) + c -c

-2 + 8 = 49a - 25a - 7b + 5b

6 = 24a - 2b

Substitutes b = 6a

6 = 24a - 2(6a)

6 = 26a - 12a

6 = 12a

a = 6/12 = 1/2

then b = 6a = 6*1/2 = 3

Finally, per the original 2nd and 3rd equations:

-2 = 49 ( 1/2) - 7(3) + c

-2 = 49/2 - 21 + c

-2 = 49/2 - 42/2 + c

-2 = 7/2 + c

-4/2 = 7/2 + c

c = -11/2

AND

-8 = 25a - 5b + c

= 25(1/2) - 5(3) + c

= 25/2 - 15 + c

= 25/2 - 30/2 + c

= -5/2 + c

so -16/2= -5/2 + c

c= -11/2

which verifies the solution to the system

The parabola is:

y = (1/2) x^2 + 3x - 11/2

= (1/2) (x^2 + 6x - 11)

= (1/2) (x^2 + 6x + 9 - 11 - 9) <-- completes the square

= (1/2) [(x^2 + 6x + 9) - 20 ]

= (1/2) [ (x+3)^2 - 20]

= (1/2)(x+3)^2 - 10

so the axis of symmetry is in fact x=-3

when x=-7, y = (1/2)(-7 +3)^2 - 10

= (1/2) (-4)^2 - 10

= (1/2)(16) - 10

= 8-10

= -2

so (-7,-2) is verified

when x = 5, y = (1/2)(-5+3)^2 - 10

= (1/2)(-2)^2 - 10

= 2-10

= -8

so (-5,-8) is verified

y = (1/2) x^2 + 3x - 11/2 = (1/2)(x+3)^2 - 10

is the official answer