Best Answer:
y = ax² + bx + c ← this is the quadratic function

y' = 2ax + b ← this is the derivative, i.e. the slope of the tangent line to the curve

The curve touches x-axis at 4 → point (4 ; 0)

y = ax² + bx + c → when: x = 4, then: y = 0

16a + 4b + c = 0

c = - 16a - 4b

The curve touches x-axis at 4 → y'(4) = 0

y' = 2ax + b → when: x = 4, then: y' = 0

8a + b = 0

b = - 8a

The curve passes through (2 ; 12)

y = ax² + bx + c → when: x = 2, then: y = 12

4a + 2b + c = 12 → recall: c = - 16a - 4b

4a + 2b - 16a - 4b = 12

- 12a - 2b = 12

- 6a - b = 6 ← recall: b = - 8a

- 6a - (- 8a) = 6

- 6a + 8a = 6

2a = 6

→ a = 3

Recall: b = - 8a

→ b = - 24

Recall: c = - 16a - 4b

c = - (16 * 3) - (4 * - 24)

c = - 48 + 96

→ c = 48

y = ax² + bx + c → you substitute by the previous values

y = 3x² - 24x + 48 ← this is the required function

y = ax² + bx + c ← this is the quadratic function

y' = 2ax + b ← this is the derivative, i.e. the slope of the tangent line to the curve

Vertex at the point (- 4 ; 1)

y = ax² + bx + c → when: x = - 4, then: y = 1

16a - 4b + c = 1

c = 1 - 16a + 4b

The curve passes through (1 ; 11)

y = ax² + bx + c → when: x = 1, then: y = 11

a + b + c = 11 → recall: c = 1 - 16a + 4b

a + b + 1 - 16a + 4b = 11

- 15a + 5b = 10

- 3a + b = 2

b = 3a + 2

Vertex at the point (- 4 ; 1) → y'(- 4) = 0

y' = 2ax + b → when: x = - 4, then: y' = 0

- 8a + b = 0 → recall: b = 3a + 2

- 8a + 3a + 2 = 0

- 5a + 2 = 0

- 5a = - 2

→ a = 2/5

Recall: b = 3a + 2

b = 3.(2/5) + 2

b = (6/5) + (10/5)

→ b = 16/5

Recall: c = 1 - 16a + 4b

c = 1 - 16.(2/5) + 4.(16/5)

c = (5/5) - (32/5) + (64/5)

c = (5 - 32 + 64)/5

→ c = 37/5

y = ax² + bx + c → you substitute by the previous values

y = (2/5).x² - (16/5).x + (37/5)

y = (2x² - 16x + 37)/5 ← this is the required function

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