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

neela | Student

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.

giorgiana1976 | Student

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:



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



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)