QUESTION 9
The given function has x-intercepts at;
[tex]x=-2[/tex] with multiplicity 1.
[tex]x=0[/tex] with multiplicity even, say 2.
[tex]x=3[/tex] with multiplicity 1.
By the factor theorem; [tex]x+2,(x-0)^2,x-3[/tex] are factors of the polynomial function.
The possible formula for the graph is
[tex]p(x)=ax^2(x+2)(x-3)[/tex]
The point (-1,4) lies on this graph
[tex]4=a(-1)^2(-1+2)(-1-3)[/tex]
[tex]4=-4a[/tex]
[tex]a=-1[/tex]
Hence a possible formula is
[tex]p(x)=-x^2(x+2)(x-3)[/tex]
QUESTION 10
The given polynomial function has x-intercept at x=-2, with and odd multiplicity, say 1.
It was given that;
[tex]p(i)=0[/tex]
This implies that [tex]x=i[/tex] is a solution.
By the complex conjugate property, [tex]x=-i[/tex] is also a solution.
By the factor theorem;
[tex]P(x)=a(x+2)(x-i)(x+i)[/tex]
Apply difference of two squares and simplify to get;
[tex]P(x)=a(x+2)(x^2+1)[/tex]
The graph passes through (2,-4).
[tex]-4=a(2+2)(2^2+1)[/tex]
[tex]-4=20a[/tex]
[tex]a=-\frac{1}{5}[/tex]
A possible formula is
[tex]P(x)=-\frac{1}{5}(x+2)(x^2+1)[/tex]
