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

Accepted Solution

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


and since f(5)=0, and a is -1 we have:

thus c=25-5b.

By ii) [tex]\displaystyle{b^2-4ac=0[/tex].

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        [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]