Question

1. decide if each set of conditions is possible. if so, give an example. if not, explain why not.

a. a 4th degree polynomial with five turning points
b. a 3rd degree polynomial with one turning point
c. a 2nd degree polynomial with one turning point

Explanation:

Step1: Recall the rule for turning - points of a polynomial

The maximum number of turning - points of a polynomial of degree $n$ is $n - 1$.

Step2: Analyze part a

For a 4th - degree polynomial ($n = 4$), the maximum number of turning - points is $n-1=4 - 1=3$. Since $5>3$, a 4th - degree polynomial with five turning - points is not possible.

Step3: Analyze part b

For a 3rd - degree polynomial ($n = 3$), the maximum number of turning - points is $n - 1=3 - 1 = 2$. A 3rd - degree polynomial can have 1 turning - point. For example, the polynomial $y=x^{3}-3x$ has a turning - point. We find its derivative $y'=3x^{2}-3 = 3(x^{2}-1)=3(x + 1)(x - 1)$. Setting $y'=0$, we get $x=-1$ and $x = 1$. These are the critical points, and the function has a turning - point.

Step4: Analyze part c

For a 2nd - degree polynomial ($n = 2$), the maximum number of turning - points is $n - 1=2 - 1=1$. A quadratic function $y=ax^{2}+bx + c$ ($a
eq0$) is a 2nd - degree polynomial. Its graph is a parabola, and a parabola has exactly one turning - point. For example, $y=x^{2}$ has a turning - point at $x = 0$.

Answer:

a. Not possible because the maximum number of turning - points of a 4th - degree polynomial is 3.
b. Possible. Example: $y=x^{3}-3x$.
c. Possible. Example: $y=x^{2}$.