Tazmin Horton

2020-12-09

Determine whether the following state-ments are true and give an explanation or counterexample.
a) All polynomials are rational functions, but not all rational functions are polynomials.
b) If f is a linear polynomial, then $f×f$ is a quadratic polynomial.
c) If f and g are polynomials, then the degrees of $f×g$ and $g×f$ are equal.
d) To graph $g\left(x\right)=f\left(x+2\right)$, shift the graph of f 2 units to the right.

Ezra Herbert

Step 1
(a) Given:
All polynomials are rational functions, but not all rational functions are polynomials.
Determine whether the following state-ments are true and give an explanation or counterexample.
Step 2
Explanation- A polynomial of degree n has at most n real zeros. The degree of a polynomial function determines the end behavior of its graph. A rational function is a function of the form $f\left(x\right)=P\left(x\right)Q\left(x\right)$ or $f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$, where P(x) and Q(x) are both polynomials.
Every polynomial function is a rational with $Q\left(x\right)=4.$ A function that cannot be written in the form of a polynomial, such as , is not a rational function.
So, the given statement "All polynomials are rational functions, but not all rational functions are polynomials" is true statement.
Hence, the given statement "All polynomials are rational functions, but not all rational functions are polynomials" is true statement.

Jeffrey Jordon

Answer is given below (on video)

xleb123

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

fudzisako

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 $f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\dots +{a}_{1}x+{a}_{0}$, 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, $f\left(x\right)=\frac{1}{x}$ is a rational function but not a polynomial.
Step 2:
b) If $f$ is a linear polynomial, then $f×f$ is a quadratic polynomial.
$\mathbf{\text{True}}$. A linear polynomial is of the form $f\left(x\right)=mx+b$, where $m$ and $b$ are constants. When we multiply this linear polynomial by itself, we get $\left(mx+b\right)×\left(mx+b\right)={m}^{2}{x}^{2}+2mbx+{b}^{2}$. 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 $f×g$ and $g×f$ are equal.
$\mathbf{\text{True}}$. Let $f\left(x\right)$ have degree $m$ and $g\left(x\right)$ 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 $f×g=g×f$. Therefore, the degrees of $f×g$ and $g×f$ are equal.
Step 4:
d) To graph $g\left(x\right)=f\left(x+2\right)$, shift the graph of $f$ $2$ units to the right.
$\mathbf{\text{True}}$. Shifting the graph of a function $f\left(x\right)$ $a$ units to the right is equivalent to replacing $x$ with $\left(x-a\right)$ in the function. In this case, $g\left(x\right)=f\left(x+2\right)$, so to graph $g\left(x\right)$, we need to shift the graph of $f\left(x\right)$ $2$ units to the right by replacing $x$ with $\left(x+2\right)$.

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