Remember that
A quartic function is a fourth-degree polynomial
so
the polynomial f(x) is of the form
f(x)=a(x-x1)(x-x2)(x-x3)(x-x4)
where
a is the leading coefficient
x1,x2,x3, and x4 are the roots of the polynomial
Looking at the given table of values
we have
x1=-2
x2=1
x3=2
x4=3
substitute
f(x)=a(x+2)(x-1)(x-2)(x-3)
Find out the value of a
For x=0, f(x)=-48
substitute
-48=a(0+2)(0-1)(0-2)(0-3)
-48=a(2)(-1)(-2)(-3)
-48=a(-12)
a=4
therefore
f(x)=4(x+2)(x-1)(x-2)(x-3)
Convert to standard form
we have that
(x+2)(x-1)=x^2-x+2x-2=(x^2+x-2)
(x-2)(x-3)=x^2-3x-2x+6=(x^2-5x+6)
substitute
f(x)=4(x^2+x-2)(x^2-5x+6)
(x^2+x-2)(x^2-5x+6)=x^4-5x^3+6x^2+x^3-5x^2+6x-2x^2+10x-12
Combine like terms
x^4-5x^3+6x^2+x^3-5x^2+6x-2x^2+10x-12=x^4-4x^3-x^2+16x-12
so
f(x)=4(x^4-4x^3-x^2+16x-12)
therefore
the answer is
f(x)=4x^4-16x^3-4x^2+64x-48