Answer
a. $x=-1$, odd multiplicity, $x=3$, even multiplicity
b. $y=(x+1)(x-3)^2$
c. $y=9$
Work Step by Step
Using the graph given in the exercise, we have:
a. zeros: $x=-1$, odd multiplicity, $x=3$, even multiplicity,
b. We can write the equation as $y=(x+1)(x-3)^2$
c. The y-intercept can be found as $y=9$ from the above equation by letting $x=0$, or by reading from the given graph.
