Anonymous a year ago
1.ray and kelsey are working to graph a third-degree polynomial function that represents the first pattern in the coaster plan. ray says the third-degree polynomial has 4 intercepts. kelsey argues the function can have as many as 3 zeros only. is there a way for the both of them to be correct? explain your answer

Of course there is no way for them to be both correct, since they contradict each other. Here is how to prove Ray incorrect. Suppose that a polynomial has four roots: s, t, u, and v. If the polynomial were evaluated at any of these values, it would have to be zero. Therefore, the polynomial can be written in this form. p(x)(x - s)(x - t)(x - u)(x - v), where p(x) is some non-zero polynomial This polynomial has a degree of at least 4. It therefore cannot be cubic. Now prove Kelsey correct. We have already proved that there can be no more than three roots. To prove that a cubic polynomial with three roots is possible, all we have to do is offer a single example of that. This one will do. (x - 1)(x - 2)(x - 3) This is a cubic polynomial with three roots, and four or more roots are not possible for a cubic polynomial. Kelsey is correct. Incidentally, if this is a roller coaster we are discussing, then a cubic polynomial is not such a good idea, either for a vertical curve or a horizontal curve.

Here is my answer, i'm sorry if you didn't understand my explanation