Explanation:
yes, for a function to have 3 solutions for y = 0, we need an expression in x with highest exponent 3.
such an expression can then be formed into a series of multiplications of (x + #) terms. each # is different, as these are the x values leading to y = 0 (remember, when one term in a multiplication is 0, then the whole result is 0).
f(x)=1/# × (x+#)(x+#)(x+#)
the graph tells us that y = 0 for x = -4, -1, 3
each (x+#) term must now handle one of these cases.
how is (x+#)=0, if x = -4 ? # = 4
how is (x+#)=0, if x = -1 ? # = 1
how is (x+#)=0, if x = 3 ? # = -3
so we get for the moment
f(x) = 1/# × (x+4)(x+1)(x-3)
to find the last # of 1/# we use additional point of y-axis intercept (0, -6) :
-6 = 1/# × (0+4)(0+1)(0-3) = 1/# × 4×1×-3 = 1/# × -12
-6# = -12
# = -12/-6 = 2
so, the full function is
f(x) = 1/2 × (x+4)(x+1)(x-3)