Quadratic function f

Question

Determine the quadratic function f whose graph is given.

The vertex is (4,-4) and the other given point is (2,4).

Answer f(x)=

Answer 1

y = C(x-4)^2 - 4 <--- given the vertex is (4,-4)

4 = C(2-4)^2 - 4

4 = c(-2)^2 - 4

4 = 4c - 4

8 = 4c

c = 2

y = 2(x-4)^2 - 4

= 2(x^2 - 8x + 16) - 4

= 2x^2 - 16x + 32 - 4

= 2x^2 - 16x + 28

Another way to think of it is that (6,4) is also on the parabola

y = Ax^2 + Bx+C

(4,-4) ---> -4 = 16A + 4b + C

(6,4) ---> 4 = 36A + 6B + C

(2,4) ---> 4 = 4A + 2B + C

solves 3 x 3.....

subtracts Eq2 - Eq1: 8 = 20a + 2b

4 = 10a + b

subtracts eq2 - eq3: 0=32a + 4b

0 = 8a + b

b = -8a

4 = 10a + -8a

4 = 2a

a = 2

b=-16

4 = 4A + 2B + C

4 = 4(2) + 2(-16) + c

4 = 8 - 32 + c

4 = -24 + c

c = 28

y = 2x^2-16x + 28 which agrees with the previous answer..