Answer
a. zeros: -2, 1, and 4
b. $f(x)=(x+2)(x-1)(x-4)$
c. y-intercept: 8
Work Step by Step
a. Zeros: -2, 1 and 4 (x-intercepts of the graph)
At all the zeros, f(x) crosses the x-axis,
all zeros have odd multiplicity.
b. Factors of f(x): $(x+2),\ (x-1)$, and $(x-4)$
There are two turning points, so the degree is at least 3.
To make the degree the lowest possible we take multiplicity of 1 for all three zeros.
This will make the degree of f be 3.
For an odd degree, positive leading coeddicient
we have end behaviour $(\swarrow,\nearrow)$,
so we take the leading coefficient to be 1
$f(x)=(x+2)(x-1)(x-4)$
c. Reading the graph, the y-intercept seems to be 8
$f(0)=(0+2)(0-1)(0-4)=8$
