Find the triplet (a,b,c) for f(x)=ax^2+bx+c, if f is canceling for x=1 and f has a minimal point=-25/12, when x=1/6
By remainder theorem, if x-a is a factor o f(x) then f(a) = 0 or a*^2+b*1+c= o Or a+b+c=0 (1).
f(1/6) = -25/6 Or a(1/6)^2+b(1/6)+c= -25/6. Or multiplying by 36, we get:a+6b+36c = -150...(2).
Since f(1/6) is minimum at (1/6), f'(x) at 1/6 is zero: 2a(1/6)+b = 0. Or a = -3b.....(3).
36*(1)-(2) gives: 35a+30b = 75. Using (3) , 35(-3b)+30b = 75. Or -75b = 75. Or b = -1. So from (3) , a = -3b = 3. and from (1) 3-1+c = 0. Or c = -2.
In order to find out the coefficients a,b,c, we have to have 3 relationships, which we could write from the data given by the enunciation.
If f is zero for x=1, we could write:
Another clue given by enunciation is that we have the coordinates of the minimal point (1/6, -25/12).
But we know that the coordinates of the minimal point are
x=-b/2a, but x=1/6
y=-delta/4a, but y= -25/12, where delta=b^2-4ac
We'll substitute -3b=a into the relation a+b+c=0
Into the relation 25a=3(b^2-4ac), we'll substitute -3b=a and c=2b, so that we'll obtain an equation with the single unknown, which is b.
After reducing similar terms, we'll have:
b1=0 and b2=-1
We'll choose the value for b, so that the value of a to be positive, a>o, so that we could have a minimal point.
For b=0, -3b=a, a=0 and this value is not respecting the constraint a>0.
For b=-1, a=3>0 and c=2b, so c=-2.
So, the triplet which is respecting all function constraints is: (3,-1,-2)