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Sedbober [7]
3 years ago
9

Numer 9 and 10. Need Solutions​

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0

QUESTION 9

The given function has x-intercepts at;

x=-2 with multiplicity 1.

x=0 with multiplicity even, say 2.

x=3 with multiplicity 1.

By the factor theorem; x+2,(x-0)^2,x-3 are factors of the polynomial function.

The possible formula for the graph is

p(x)=ax^2(x+2)(x-3)

The point (-1,4) lies on this graph

4=a(-1)^2(-1+2)(-1-3)

4=-4a

a=-1

Hence a possible formula is

p(x)=-x^2(x+2)(x-3)

QUESTION 10

The given polynomial function has x-intercept at x=-2, with and odd multiplicity, say 1.

It was given that;

p(i)=0

This implies that x=i is a solution.

By the complex conjugate property, x=-i is also a solution.

By the factor theorem;

P(x)=a(x+2)(x-i)(x+i)

Apply difference of two squares and simplify to get;

P(x)=a(x+2)(x^2+1)

The graph passes  through (2,-4).

-4=a(2+2)(2^2+1)

-4=20a

a=-\frac{1}{5}

A possible formula is

P(x)=-\frac{1}{5}(x+2)(x^2+1)

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