How can you solve this? Not only does it say to write THE equation ..., it says to write THE EXACT equation.

[Steven G]

Not only does it say to write THE equation ..., it says to write THE EXACT equation.
I see four unknowns that need to be solved for but see only two or three points on the graph other than the zeros. Sure you can tell that the leading coefficient is positive from end behavior but from what I know there are quite a few positive numbers.
I got as far as f(x)=A(x+3)^2b+1(x+1)^2c+1(x-2)^2d

 

[Dr.Peterson]

Not only does it say to write THE equation ..., it says to write THE EXACT equation.
I see four unknowns that need to be solved for but see only two or three points on the graph other than the zeros. Sure you can tell that the leading coefficient is positive from end behavior but from what I know there are quite a few positive numbers.
I got as far as f(x)=A(x+3)^2b+1(x+1)^2c+1(x-2)^2d
View attachment 35888

It's clear that the zero at -3 has multiplicity 1, not some other odd number, because it is not horizontal.

The others are clearly intended to be multiplicity 3 and 2, because higher multiplicities would be flatter. So although you are correct that, in principle, the latter could be any odd and even multiplicity, it is reasonable to assume 3 and 2. And when you then use the y-intercept to determine A, you can check the local max and min and confirm that guess.

It does pass the check:

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[Dr.Peterson]

For comparison, if you change the multiplicities to 5 and 4, here is what you get (in blue):

​

 

[Steven G]

It's clear that the zero at -3 has multiplicity 1, not some other odd number, because it is not horizontal

​

Interesting.

 

[Dr.Peterson]

A general idea I teach is that, near a zero of multiplicity n, the graph looks like the polynomial [imath]x^n[/imath].

More specifically, if the zero is at c, the graph of [imath]A(x-c)^n[/imath] approximates the curve, where A is obtained by plugging c into the other factors. In this case, for example, my polynomial is [imath]\left(x+3\right)\left(x+1\right)^{3}\left(x-2\right)^{2}[/imath], so near [imath]x=-1[/imath], the graph is close to [imath]y=\left(-1+3\right)\left(x+1\right)^{3}\left(-1-2\right)^{2} = 18\left(x+1\right)^{3}[/imath], like this:

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So near [imath]x=-3[/imath], with multiplicity 1, it looks like a straight line.