Data:
x — intercept of multiplicity 2=1.3

© — intercept of multiplicity 1=-3

y—intercept = 9

Degree=5

Since it is a fifth degree polynomial function with multiplicity of 2 and 1 for some zeros, its general equation becomes: \(\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}+{3}\right)}{\left({x}—{1}\right)}^{{2}}{\left({x}-{3}\right)}^{{2}}\)

In order to evaluate a, use the y - intercept (0,9), therefore substitute f(0)=9 in this equation:

\(\displaystyle{9}={a}{\left({0}+{3}\right)}{\left({0}—{1}\right)}^{{2}}{\left({0}-{3}\right)}^{{2}}\)

Simplify: 9=27a

Evaluate a: \(\displaystyle{a}=\frac{{9}}{{27}}=\frac{{1}}{{3}}\)

This implies that the equation of the given polynomial function is f(x) =

\(\displaystyle\frac{{1}}{{3}}{\left({x}+{3}\right)}{\left({x}—{1}\right)}^{{2}}{\left({x}-{3}\right)}^{{2}}\)

© — intercept of multiplicity 1=-3

y—intercept = 9

Degree=5

Since it is a fifth degree polynomial function with multiplicity of 2 and 1 for some zeros, its general equation becomes: \(\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}+{3}\right)}{\left({x}—{1}\right)}^{{2}}{\left({x}-{3}\right)}^{{2}}\)

In order to evaluate a, use the y - intercept (0,9), therefore substitute f(0)=9 in this equation:

\(\displaystyle{9}={a}{\left({0}+{3}\right)}{\left({0}—{1}\right)}^{{2}}{\left({0}-{3}\right)}^{{2}}\)

Simplify: 9=27a

Evaluate a: \(\displaystyle{a}=\frac{{9}}{{27}}=\frac{{1}}{{3}}\)

This implies that the equation of the given polynomial function is f(x) =

\(\displaystyle\frac{{1}}{{3}}{\left({x}+{3}\right)}{\left({x}—{1}\right)}^{{2}}{\left({x}-{3}\right)}^{{2}}\)