Question

Give a possible formula of minimum degree for the polynomial h(x) in the graph below?

Answer 1

`h(x) =1/15 (x +3)(x-3)(x+1)(x-1)^2`

We got the factors from the zeros. It's odd degree since it has both `+infty` and `-infty` as ends. `x=1` is tangent, so we squared `x-1.` At `x=2` we have `(2-3)(2+3)(2+1)(2-1)^2=-15` and we want `h(2)=-1` so we multiply by `1/15.`

Explanation:

graph{1/15 (x +3)(x-3)(x+1)(x-1)^2 [-10, 10, -5, 5]}