Which second-degree polynomial function has a leading coefficient of –1 and root 5 with multiplicity 2?
Question
Answer:
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]
solved
general
10 months ago
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