Since you are dealing with functions and relations, polynomials are probably going to be some of the major concerns. Here are a few important details you may need to know about them.

## What are polynomials?

A polynomial is known to be a series of terms. Also, each one of them is supposed to be the product of a constant coefficient and an independent variable with integer power. A general polynomial function *f *has been expressed below with respect to a variable *x*.

f(x) = a_{n }x^{n} + a_{n −1} x^{n−1} + . . . + a_{2}x^{2} + a_{1}x + a_{0 }

Here, the highest power of x is said to be the degree of the given polynomial. Constant polynomials hold on to the degree zero, linear polynomials have it as 1, quadratic polynomials 2, cubics 3 and quartics as 4. The function f(x) = 0 is a polynomial. However, its degree is said to be undefined.

## Finding the domain and range of a polynomial

The domain and range for a function could be found graphically or without graph. Consider finding the domain and range for the given function $f(x)=\frac{\sqrt{x+2}}{x^{2}-9}$

Take a look at the numerator (i.e. the top of the fraction), it is a square root. In order to make sure that the root for the same is positive we will have to consider the x values that are either equal to or greater than 2.

The denominator or the bottom of the fraction has$ x^{2}-9$. It could be also written as (x+3) (x-3). Therefore, the values for x cannot include -3 or 3 from the first or second bracket. -3 can however be ignored since we know that $x\ge $ -2. The domain in this case, would, therefore, be $x\ge $ -2 or $x\ne $ 3.

In order to make out the range, we will have to consider the numerator and denominator separately. In the numerator, if$ x=-2$, then the numerator would further be considered as $\sqrt{2+2}$ which would be $\sqrt{0}$ =0. As the value of$x$ increases from -2, the numerator also tends to increase.

Now, breaking down the denominator, when $x=-2$, it will be (-2)$^{2}$ – 9 = 4 – 9 = -5. We have,$ f(-2)=\frac{0}{-5}=0$. Among $x=-2$ and $x=3$, $ (x^{2}-9)$ is closer to zero. In that case, $f(x)$ tends to get closer to $-\infty $ as it gets even nearer to $x=3. $

When $x>3$, i.e. $x$ is larger than 3, the value of the denominator is likely to be just over zero. Hence, $f(x)$ is supposed to be a large number that is positive.

The range of the given function can therefore concluded to be $(-\infty ,0\rbrack \cup (\infty ,0. $