Question

find the parabola whose minimum is at (7,6) rather than the point given in the book.
The parabola's equation is y = x2 + ax + b, where
a = and
b =

Answer 1

(dy/dx)=2x+a
Funcktion have extreme value where is
(dy/dx)=0
2x+a=0 ,2x=-a, Divided with 2,
x=-(a/2)
For x=7 , 7=(-a/2) , Multipled with 2
7*2=-a , 14=-a , Multilpled with (-1)
a=-14

y=x^2+ax+b

x=7 y=6 a=-14

6=(-a/2)^2+a*(-a/2)+b
6=(14/2)^2-14*(-14/2)+b
6=7^12-14*7+b
6=49-98+b , 6=-49+b , b=6+49 , b=55
y=x^2+ax+b
y=x^2-14x+55
For x=7
y=7^2-14*7+55=49-98+55=6

Answer 2

Proof that is minimum.
Function have minimum if (dy/dx)=0
and (d^2y/dx^2)>0
If second derivate greater than zero.
Second derivate is derivate of first derivate.
In this case:
(d^2y/dx^2)=d(2x)/dx=2>0
That is minimum