The curve above is the graph of a degree 3 polynomial. it goes through the point (5, -21.6). find the polynomial.

We can use what we know about multiplicities and standard forms of graphs to solve without knowing that point.

3rd degree normal form is f(x)=ax^3+bx^2+cx+d
We know that a 3rd degree function with a positive leading coefficient (a) will go from bottom left to top right.
With a negative leading coefficient, it goes from top left to bottom right.

We know that a is negative.

multiplicities
When a point goes through, it has odd multiplicity, in this case, 1
When it is is tangent, it has an even multiplicity or in this case, 2

We see the intercepts are at x=-3 and at x=2, the x=-3 passes through 1 time but at x=2, it is tangent or bounces off the x axis.
with roots r₁ and r₂, the equation is (x-r₁)(x-r₂)

the roots are x=-3 and x=2, but x=2 appears 2 times so
(x-(-3))(x-2)^2=(x+3)(x-2)(x-2)

remember that it has negative leading coefficient
f(x)=-(x+3)(x-2)(x-2)
f(x)=(3-x)(x-2)(x-2)
or expanded
f(x)=-x³+x²+8x-12