Find constants - Mathematics?

3x² - 5x + 1 = a(x+b)² + c

3x² - 5x + 1 = ax² + 2ab.x + a.b² + c

this is only possible if all powers of x vanish so:

a = 3

this gives

- 5x + 1 = + 6b.x + 3.b² + c

so b = -5/6

this gives

+ 1 = + 25/12 + c

so c = -13/12

You can find the vertex (lowest value) of y = 3x² - 5x + 1 by differentiating

dy/dx = 0 = 6.x - 5 so at x = 5/6

y = 3(x^2 - 5/3*x + (5/6)^2) + 1 - 25/36

y = 3(x - 5/6)^2 - 13/12 => in vertex form of : a(x - h)^2 + k, where (h , k) is the vertex:

vertex at(5/6 , -13/12)

y = 3x² - 5x + 1

you want to convert it to "vertex form," y = a(x - h)² + k

factor out the leading coefficient

y = 3(x² -(5/3)x) + 1

complete the square

 coefficient of x term: -5/3

 divide coefficient in half: -5/6

 square it: (5/6)²

 use (5/6)² to complete the square:

y = 3(x² -(5/3)x + (5/6)²) - 3(5/6)² + 1

 = 3(x - 5/6)² - 3(5/6)² + 1

 = 3(x - 5/6)² - 13/12

vertex (5/6, -13/12)