This is a practice problem from precalculus?

With point (0,0), you will know that

0 =a(0)^2+b(0)+c --> c=0

Put (3,-3) and (6,0) into the equation you get

-3 =a(3)^2+b(3) --> -3 = 9a + 3b

0=a(6)^2+b(6) --> 0 = 36a + 6b --> b = -6a

Put b = -6a back into -3 = 9a + 3b, you get

-3 = 9a + 3(-6a) --> -3 = -9a --> a = 1/3

Since b = -6a, then b = -6(1/3) = -2

SO your answers will be y = (1/3)x^2 - 2x

The coordinates of all three points must

satisfy the equation y = ax² + bx + c and,

since the curve passes through the origin,

c = 0

For (3, - 3):

x = 3

y = - 3

- 3 = a(3)² + b(3) + 0

- 3 = a(9) + 3b

- 3 = 9a + 3b

9a + 3b = - 3

3a + b = - 1

b = - 3a - 1 ......................... Eq. 1

For (6, 0):

x = 6

y = 0

0 = a(6)² + b(6)

0 = a(36) + 6b

0 = 36a + 6b

36a + 6b = 0

6a + b = 0 ......................... Eq. 2

Subbing (- 3a - 1) from Eq. 1 for b in Eq. 2,

6a + (- 3a - 1) = 0

6a - 3a - 1 = 0

3a = 1

a = 1/3

Subbing 1/3 for a in Eq. 1,

b = - 3(1/3) - 1

b = - 1 - 1

b = - 2

Equation:

y = 1/3 x² - 2x

¯¯¯¯¯¯¯¯¯¯¯

 

(0,0) (3,-3), (6,0)

vertex is (3, -3) ... 3 = -b/2a

so b = -6a

(0,0) ==>> 0 = a(0) +b(0) + c ==>> c = 0

y = ax^2 + bx + c

(3, -3) ==>> -3 = 9a + 3b ... multiply by -2

(6, 0) ==>> 0 = 36a + 6b

6 = -18a - 6b

0 = 36a + 6b ... add

6 = 18a

a = 1/3 ... b = -6a = -2 and c = 0

y = (1/3)x^2 - 2x << answer

Sub in those points:

0 = 0a + 0b + c

Therefore, c = 0

[1] -3 = 9a + 3b (+ c, which is 0)

[2] 0 = 36a + 6b

Let [3] = [2] - 2[1]

[3] 6 = 18a

a = ⅓

[2] 0 = 36(⅓) + 6b

-12 = 6b

b = -2

Therefore, the equation is y = ⅓x² - 2x

c=0

y=ax(x-6)

-3=a(3)(3-6)

-3=-9a

1/3=a

y=(x^2-6x)/3