Question

find the constants a, b and c such that or all values of x,

3x^2 - 5x + 1 = a(x+b)^2 +c

also whats the coordinate of the minimum point??

Answer 1

3x^2-5*x+1=a*(x^2+2*b*x+b^2)+c

Becouse):
(x+b)^2=x^2+2*b*x+b^2

3x^2-5*x+1=a*x^2+2*a*b*x+a*b^2+c
3x^2-5*x+1=
=a*x^2+(2*a*b)*x+(a*b^2+c)

a*x^2=3*x^2 Divide with x^2

a=3

(2*a*b)*x=-5x Divide with x
2*a*b=-5
2*3*b=-5
6b=-5 Divide with 6

b=(-5/6)

a*b^2+c=1
3*(-5/6)^2+c=1
(3*25/36)+c=1
(25/12)+c=1
(25/12)+c=1
c=1-(25/12)=(12/12)-(25/12)=(-13/12)

c=(-13/12)

a=3 , b=(-5/6) , c=(-13/12)

Function have extreme where first derivate (dy/dx)=0
d(3x^2-5*x+1)/dx
=2*3*x-5
=6*x-5
(dy/dx)=6*x-5
6*x-5=0
6*x=5 Divide with 6

x=(5/6)
y=3*(5/6)^2-5*(5/6)+1
=3*(25/36)-(25/6)-1
=(25/12)-(25/6)-1
=(25/12)-(50/12)-(12/12)=(-37/12)

Coordinates of extreme point:
(5/6 , -37/12)

Extreme point is minimum if secod derivate >0

Secon derivate (d^2y/dx^2) is derivate of first derivate.
In this case:
(d^2y/dx^2)=6>0
That is minimum.