A second degree polynomial function has the general form:
, where 
.
The leading coefficient is a, so we have a=-1.
5 is a double root means that :
i) f(5)=0,
ii) the discriminant D is 0, where
.
Substituting x=5, we have
f(5)=a(5)^2+b(5)+c,
and since f(5)=0, and a is -1 we have:
0=-25+5b+c
thus c=25-5b.
By ii)
.
Substituting a with -1 and c with 25-5b we have:
Finally we find c: c=25-5b=25-50=-25
Thus the function is
Remark: It is also possible to solve the problem by considering the form
directly.
In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is
The leading coefficient is a, so we have a=-1.
5 is a double root means that :
i) f(5)=0,
ii) the discriminant D is 0, where
Substituting x=5, we have
f(5)=a(5)^2+b(5)+c,
and since f(5)=0, and a is -1 we have:
0=-25+5b+c
thus c=25-5b.
By ii)
Substituting a with -1 and c with 25-5b we have:
Finally we find c: c=25-5b=25-50=-25
Thus the function is
Remark: It is also possible to solve the problem by considering the form
In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is