# 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

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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:

f(1)=0

f(1)=a+b+c

So, a+b+c=0

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

V(-b/2a, -delta/4a)

x=-b/2a, but x=1/6

So, -b/2a=1/6

-6b=2a

-3b=a

y=-delta/4a, but y= -25/12, where delta=b^2-4ac

(b^2-4ac)/4a=25/12

25*4a=12(b^2-4ac)

25a=3(b^2-4ac)

We'll substitute -3b=a into the relation a+b+c=0

-3b+b+c=0

-2b=-c

c=2b

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.

25(-3b)=3[b^2-4(-3b)(2b)]

After reducing similar terms, we'll have:

-25b=b^2+24b^2

-25b=25b^2

b^2+b=0

b(b+1)=0

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)**