Determine whether the following state-ments are true and give an explanation or counterexample. a) All polynomials are rational functions, but not all

a) The statement is true. A polynomial is a function of the form $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_{1}x+a_0$, where $a_n,a_{n-1},\dots ,a_1,a_0$ are constants and $n$ is a non-negative integer. A rational function, on the other hand, is a function of the form $R(x)=\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials and $Q(x)\ne 0$.
All polynomials can be expressed as rational functions by setting the denominator $Q(x)$ to be the constant polynomial $1$. For example, the polynomial $P(x)=2x^2-3x+1$ can be written as the rational function $R(x)=\frac{2x^2-3x+1}{1}$.
However, not all rational functions are polynomials. For instance, consider the rational function $R(x)=\frac{1}{x}$. The function $R(x)$ is not a polynomial because it has a non-integer power of $x$ in the denominator.
b) The statement is true. A linear polynomial is of the form $f(x)=mx+b$, where $m$ and $b$ are constants. When we multiply $f(x)$ by itself, we get:
$f(x)\times f(x)=(mx+b)\times (mx+b)=m^2{x}^2+2mbx+b^2$
The resulting expression is a quadratic polynomial. Therefore, if $f(x)$ is a linear polynomial, $f(x)\times f(x)$ is a quadratic polynomial.
c) The statement is true. Let $f(x)$ and $g(x)$ be polynomials of degrees $n$ and $m$ respectively. The degree of a polynomial is determined by the highest power of $x$ in the polynomial.
When we multiply $f(x)$ and $g(x)$, the resulting polynomial will have terms obtained by multiplying the terms of $f(x)$ and $g(x)$. The highest power of $x$ in the resulting polynomial will be the sum of the powers of $x$ in the terms with the highest powers from $f(x)$ and $g(x)$.
Therefore, the degree of $f(x)\times g(x)$ will be $n+m$. Similarly, the degree of $g(x)\times f(x)$ will also be $n+m$. Hence, the degrees of $f(x)\times g(x)$ and $g(x)\times f(x)$ are equal.
d) The statement is true. To graph the function $g(x)=f(x+2)$, we need to shift the graph of $f(x)$ two units to the right. This means that for each point $(x,y)$ on the graph of $f(x)$, we need to plot the corresponding point $(x-2,y)$ on the graph of $g(x)$.
For example, if the point $(1,3)$ is on the graph of $f(x)$, then the corresponding point on the graph of $g(x)$ would be $(1-2,3)=(-1,3)$.
By shifting the graph of $f(x)$ two units to the right, we obtain the graph of $g(x)$.

Step 1:
a) All polynomials are rational functions, but not all rational functions are polynomials.
$\mathbf{\text{True}}$. A polynomial is defined as a function of the form $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_{1}x+a_0$, where $a_i$ are coefficients and $n$ is a non-negative integer. A rational function, on the other hand, is defined as the quotient of two polynomials, where the denominator is not the zero polynomial. Every polynomial can be expressed as a rational function by considering it as the quotient of itself and the constant polynomial $1$. However, not all rational functions can be expressed as polynomials. For example, $f(x)=\frac{1}{x}$ is a rational function but not a polynomial.
Step 2:
b) If $f$ is a linear polynomial, then $f\times f$ is a quadratic polynomial.
$\mathbf{\text{True}}$. A linear polynomial is of the form $f(x)=mx+b$, where $m$ and $b$ are constants. When we multiply this linear polynomial by itself, we get $(mx+b)\times (mx+b)=m^2{x}^2+2mbx+b^2$. This is a quadratic polynomial since the highest power of $x$ is $2$.
Step 3:
c) If $f$ and $g$ are polynomials, then the degrees of $f\times g$ and $g\times f$ are equal.
$\mathbf{\text{True}}$. Let $f(x)$ have degree $m$ and $g(x)$ have degree $n$. When we multiply these polynomials, the degree of the resulting polynomial will be $m+n$. However, multiplication of polynomials is commutative, so $f\times g=g\times f$. Therefore, the degrees of $f\times g$ and $g\times f$ are equal.
Step 4:
d) To graph $g(x)=f(x+2)$, shift the graph of $f$ $2$ units to the right.
$\mathbf{\text{True}}$. Shifting the graph of a function $f(x)$ $a$ units to the right is equivalent to replacing $x$ with $(x-a)$ in the function. In this case, $g(x)=f(x+2)$, so to graph $g(x)$, we need to shift the graph of $f(x)$ $2$ units to the right by replacing $x$ with $(x+2)$.