Given the quadratic function y=x^2+bx+c, find the values of the constants b and c so that 1 becomes the minimum value at x=2?

The coefficient of x² term is 1.
The quadratic function that minimum y = 1 at x = 2 :
y = (x - 2)² + 1 ...... [1]

The above be function can be expressed as :
y = x² + bx + c ...... [2]

[2] = [1] :
x² + bx + c = (x - 2)² + 1
x² + bx + c = x² - 4x + 4 + 1
x² + bx + c = x² - 4x + 5

Compare the coefficients :
b = -4
c = 5

y = x^2 + bx + c

y ' = 2x + b

For min 2x + b = 0

x = - b/2

At x = 2

2 = - b/2

b = - 4

For x = 2, b = -4 and y = 1

1 = 2^2 + (-4)(2) + c

1 = 4 - 8 + c

c = 5

Differentiate
dy/dx = 2x + b = 0 ( Min value ; horizontal gradient).
At x = 2
2(2) = -b
4 = -b
b = -4

Hence
y = x^2 - 4x + c
When y = 1
1 = x^2 - 4x + c
1 = 2^2 - 4(2) + c
1 = 4 - 8 + c
c = 5
Hence
y = x^2 - 4x + 5

y = x² + bx + c

plug in (2,1)
1 = 2² + 2b + c
-3 = 2b+c
c = -2b - 3

dy/dx = 2x+b = 0
2·2 + b = 0
b = -4
c = -2(-4) - 3 = 5

y = x² - 4x + 5

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