Answer:
(x-2),(x-1),(x), and(x+1)
Step-by-step explanation:
In this scenario, we are given nothing about the polynomial except its graph. We can start in problems like these by looking at when y=0 and going from there.
In this graph, we can see that y=0 when x=2, 1, 0, and -1. Looking at our options, for y to be 0, there must be a factor where, when we plug x=2 in, y=0. The only option here is (x-2). Similarly, for x=1, we have (x-1), for x=0, we have x, and for x=-1, we have (x+1). The polynomial has 3 turning points, so the polynomial must be of degree 4 or higher.
So far, we have (x-2)(x-1)(x)(x+1). Plugging this in to a graphing calculator, we can see that this fits. To check if x²+2 and x²+1 would work here, it would require guess and check with values/graphing calculators, seeing if they fit. For example, when x=1.5, y is greater than -1 on the graph and our function. However, if we multiplied that by x²+1 or x²+2. when x=1.5, y would balloon downward, and the graph wouldn't fit
