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

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)