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Valentin [98]
3 years ago
14

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

Mathematics
2 answers:
Julli [10]3 years ago
6 0

Answer:

f(x) = 2x2 – 4x – 30

Step-by-step explanation:

Got it right

PIT_PIT [208]3 years ago
5 0
A second degree polynomial function has the general form:

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

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 D=b^2-4ac.

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) \displaystyle{b^2-4ac=0.

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


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

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


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

f(x)=-1(x-5)^2 directly.

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