Question

1. circle the following that are examples of polynomial functions. y = 4π y=-3x^(-2) - 1 y=∛(-8x^2) y=√(6x) y=(5 - x)(x - 5)^3 y=√(-3x) y=(x^2 + 8x)/x y=√(9x^2 - 1) y=7x - 2x^(-1)

Explanation:

Step1: Recall polynomial function definition

A polynomial function in one - variable $x$ is of the form $y = a_nx^n+a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0$, where $n$ is a non - negative integer and $a_i$ are real numbers.

Step2: Analyze $y = 4\pi$

It is a constant function. A constant function $y = c$ (where $c$ is a real number) is a polynomial function of degree 0. Here $a_0=4\pi$ and $n = 0$.

Step3: Analyze $y=(5 - x)(x - 5)^3$

Expand $(5 - x)(x - 5)^3=-(x - 5)(x - 5)^3=-(x - 5)^4$. Using the binomial theorem $(a - b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}(-b)^k$, we can expand it into the form of $a_nx^n+a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0$, so it is a polynomial function.

Step4: Analyze $y = 7x-2x^{-1}$

Since it has a term with a negative exponent ($x^{-1}$), it is not a polynomial function.

Step5: Analyze $y=\frac{x^2 + 8x}{x}=x + 8$ ($x

eq0$)
The original form has a variable in the denominator. In the strict sense of a polynomial function (defined for all real $x$), the function $y=\frac{x^2+8x}{x}$ is not a polynomial function because its domain is $\mathbb{R}\setminus\{0\}$, while a polynomial function is defined for all real $x$.

Step6: Analyze $y=\sqrt{9x^2-1}$

It is a radical function, not a polynomial function as it cannot be written in the form $a_nx^n+a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0$.

Step7: Analyze $y=\sqrt{6x}$

It is a radical function, not a polynomial function.

Step8: Analyze $y=\sqrt{-3x}$

It is a radical function, not a polynomial function.

Step9: Analyze $y=\sqrt[3]{-8x^3}=- 2x$

It is a polynomial function of degree 1.

Step10: Analyze $y=-3x^{-\frac{1}{2}}$

It has a non - integer exponent, so it is not a polynomial function.

Answer:

$y = 4\pi$, $y=(5 - x)(x - 5)^3$, $y=\sqrt[3]{-8x^3}$