The Curves below all have equations of the form y=x^2+bx+c. In each case find the values of b and c. Show your workings?

(i)
The vertex of the parabola is (3, 1) and the coefficient of x² term is 1.
The equation of the curve: y = (x - 3)² + 1
y = x² - 6x + 9 + 1
y = x² - 6x + 10
Hence, b = -6 and c = 10

(ii)
Similarly, the equation of the curve: y = (x + 1)² - 1
y = x² + 2x + 1 - 1
y = x² + 2x
Hence, b = 2 and c = 0

(iii)
Similarly, the equation of the curve: y = (x - 4)² + 0
y = x² - 8x + 16
Hence, b = -8 and c = 16

(iv)
Similarly, the equation of the curve: y = (x + 3)² + 2
y = x² + 6x + 9 + 2
y = x² + 6x + 11
Hence, b = 6 and c = 11

y = x² + bx + c ← this is a function

y' = 2x + b ← this is its derivative


Point (3 ; 1)

y' = 2x + b → when: x = 3, then: y' = 0

6 + b = 0

→ b = - 6

y = x² + bx + c → when: x = 3, then: y = 1

9 + 3b + c = 1

c = - 8 - 3b → recall: b = - 6

→ c = 10


Point (- 1 ; - 1)

y' = 2x + b → when: x = - 1, then: y' = 0

- 2 + b = 0

→ b = 2

y = x² + bx + c → when: x = - 1, then: y = - 1

1 - b + c = - 1

c = b - 2 → recall: b = 2

→ c = 0


Point (4 ; 0)

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

8 + b = 0

→ b = - 8

y = x² + bx + c → when: x = 4, then: y = 0

16 + 4b + c = 0

c = - 4b - 16 → recall: b = - 8

→ c = 16


Point (- 3 ; 2)

y' = 2x + b → when: x = - 3, then: y' = 0

- 6 + b = 0

→ b = 6

y = x² + bx + c → when: x = - 3, then: y = 2

9 - 3b + c = 2

c = 3b - 7 → recall: b = 6

→ c = 11