Which second-degree polynomial function has a leading coefficient of –1 and root 5 with multiplicity 2?

A second degree polynomial function has the general form:

                  [tex]\displaystyle{f(x)=ax^2+bx+c[/tex], where [tex]a\neq0[/tex].

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 [tex]D=b^2-4ac[/tex].

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) [tex]\displaystyle{b^2-4ac=0[/tex].

Substituting a with -1 and c with 25-5b we have:
                                     
          [tex]\displaystyle{b^2-4ac=0[/tex]
          [tex]\displaystyle{b^2-4(-1)(25-5b)=0[/tex]
          [tex]\displaystyle{b^2+4(25-5b)=0[/tex]
          [tex]\displaystyle{b^2-20b+100=0[/tex]
          [tex]\displaystyle{(b-10)^2=0[/tex]
          [tex]\displaystyle{b=10[/tex] 


Finally we find c: c=25-5b=25-50=-25

Thus the function is        [tex]\displaystyle{f(x)=-x^2+10x-25[/tex]


Remark: It is also possible to solve the problem by considering the form

[tex]f(x)=-1(x-5)^2[/tex] directly.

In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is [tex]f(x)=a(x-r)^2[/tex]