In Precalculus, we study many topics such as sets, real numbers, complex numbers, composite and polynomial functions, trigonometry, limits, vectors, metrics, conic section, logarithm function, basic graphs, binomial theorem, mathematical induction, sequence, series, roots, rational exponents, quadratic functions, inverse functions, inequality involving rational functions, algebraic simplification and many more. Precalculus is also known as introduction to analysis. Precalculus is studied from primary schools to research schools to help students in understanding important topics that will be used to solve problems in calculus. Precalculus deals with trigonometry and explains the concepts which are introductory to calculus.

Domain of a function can be defined as all possible values of ‘$x$’ coordinate. It consists of all real numbers except those numbers that make the denominator equal to zero. Range of a function is equal to possible values of ‘$y$’ coordinates.
Suppose we have some values $(13,  56),\ ( 38, 47),\ (75,  99),\ (  65, 19)$.
Then, Domain of function is ‘$x$’ coordinate values. So, domain = $(13, 38, 75, 65)$ and range is ‘$y$’ coordinate values. So, range = $( 56, 47,  97, 19)$.
Now, we will discuss how to find Domain and Range of a Function in detail.
To find domain and range of a function we will use steps given below:
Step 1: Find the values of ‘$x$’ and ‘$y$’ coordinates of function.
Step 2: Now, values of '$x$' is domain of function and values of '$y$' is the range.
For Example: Suppose we have a function $q$ = $p^{2}  6$.
Step 1: Our function is, $q$ = $p^{2}  6$.
Step 2: Then, assume some values for ‘$p$’ coordinate to find value of ‘$q$’.
Put $p$ = $0$
$q$ = $p^{2}  6$
$q$ = $(0)^{2}  6$
$q$ = $ 6$
Put $p$ = $2$
$q$ = $p^{2}  6$
$q$ = $(2)^{2}  6$
$q$ = $4  6$
$q$ = $ 2$
Put $p$ = $5$, we get
$q$ = $p^{2}  6$
$q$ = $(5)^{2}  6$
$q$ = $25  6$
$q$ = $19$
So, value of ‘$p$’ and ‘$q$’ are:
$(0,  6),\ (2,  2),\ (5, 19)$.
Here, values of '$p$' are domain of function.
Domain = $(0, 2, 5)$
All values of '$q$' are range of function
Range = $( 6,  2, 19)$.
A function that contains two polynomials in fraction form or written in ratios is known as rational function.Rational function is of the form $f(a)$ = $\frac{R(a)}{S(a)}$
Here, both '$R$' and '$S$' are polynomial functions in '$a$' and value of '$S(a)$' is not zero. Example of rational function is $\frac{x + 7}{ x^{2}  3}$.
The exponential function is the function $e^{x}$, where $e^{x}$ will have its own derivative. When dealing with functions such as $h(x)$ = $8^{x},\ 8$ will be the base number and power is the variable $(x)$. Domain of a function or relation is defined by set of possible values on $x$axis and range are the set of possible $y$values. Let us consider the following example of an exponential function in which variable '$x$' is used as an exponent to determine domain and range of exponential Function.
Graph shown above is for the function $y$ = $2x$.
Since '$x$' can be any real number, the domain is all Real Numbers and is given as $D$ = $R:\ R$ is all real numbers.
It is clear from the figure that as '$x$' increases, curve's distance from $x$axis also increases and as '$x$' decreases, curve gets closer to $x$axis. This stops the graph at $5$. We know that range is equal to all possible $y$values. As we know that curve never goes below $x$axis, that's why $y$values are neither zero nor negative. Hence, range for exponential function will be given as all real numbers and can be represented as, Range = $R:\ R$ is all positive real numbers.
Quadratic function is a polynomial of degree $2$. The domain and range of quadratic function is all real numbers.
The standard form of quadratic equation is given as follows:
$ax^{2} + bx + c$ = $0$, where $a, b, c$ are known and '$x$' is the variable which is unknown. Remember '$a$' cannot be zero.
Solutions of some quadratic equations are not rational and we cannot find the solution by factoring. For such equations, there is an easy way to solve the solution by using quadratic formula.
The formula giving the roots of quadratic equation $ax^{2} + bx + c$ = $0$ is
$x$ = $\frac{b\pm \sqrt{b^{2}4ac}}{2a}$
Let us solve $4x^{2} + 5x + 2$ = $0$
Here, $a$ = $4,\ b$ = $5,\ c$ = $2$
Plugging the values for $a,\ b,$ and $c$ into the quadratic formula, we get
$x$ = $\frac{5\pm \sqrt{5^{2}4 \times 4 \times 2}}{2a}$
Simplify expression under the square root.
$x_{1,2}$ = $\frac{5\pm\sqrt{7}}{8}$
Solve for $x$
$x_{1}$ = $\frac{5}{8}$ + $\frac{1}{8}\sqrt{7i}$
$x_{2}$ = $\frac{5}{8}$  $\frac{1}{8}\sqrt{7i}$
A function is a relation in which input and output values are connected with a property so that, every input value has exactly one output value.
Composition of functions is a process of applying the result of one function to the other. Let $f\ :\ A\ \rightarrow B$ and $g\ :\ B\ \rightarrow C$ are two functions.
Then, the composition of $f$ and $g$, denoted by $fog$ or $fog\ (x)$, is defined as the function $fog\ :\ A\ \rightarrow C$ is given by $fog\ (x)$ = $f\ (g\ (x))$, for all $x$ belonging to $A$.
Notation used for composition is $(f o g)(x)$ = $f(g(x))$. This is read as "$f$ composed with $g$ of $x$" or "$f$ of $g$ of $x$".
Given the functions $f(x)$ = $3x$ and $g(x)$ = $x^{2} + 2$. Let us find $(f o g)(x)$ and $(g o f)(x)$ and prove that $(f o g)(x)\ \neq\ (g o f)(x)$.
$(f o g)(x)$ = $f(g(x))$ = $f(x^{2} + 2)$ = $3(x^{2} + 2)$ = $3x^{2} + 6$
$(g o f)(x)$ = $g(f(x))$
= $g(3x)$
= $(3x)^{2} + 2$
$(f o g)(x)\ \neq\ (g o f)(x)$
As $(f o g)(x)$ and $(g o f)(x)$ does not give us the same answer, composition of functions is not commutative.
Trigonometric functions are the functions of angles.They are also known as circular functions. Trigonometric functions are helpful to form a relation between the angle and sides of a right angled triangle. Most familiar trigonometric function are the sine, cosine and tangent. The trigonometric functions can be more precisely defined with the help of ratios, ratios of the two sides of a right angle triangle. Right angle triangle consists of a base, perpendicular and hypotenuse and depending on these three parameters, six trigonometric functions are there termed as sine function, cosine function, tangent function, cotangent function, secant function and cosecant function.
Domain and range of a trigonometric function are two very important terms in trigonometry. Domain of any trigonometric function can be defined as the set of values which can be substituted in any trigonometric function and range is defined as set of values which can be obtained on substituting the domain set. For different trigonometric functions, different set of domain and range are defined.
An inverse function is a function that undoes another function and goes in opposite direction. It is the relation formed when the independent variable is exchanged with the dependent variable in a given relation. The inverse of a function may not always be a function.
Let $f$ and $g$ be two functions. If $f(g(x))$ = $x$ and $g(f(x))$ = $x$, then, $g$ is the inverse of $f$ and $f$ is the inverse of $g$.
A function $f$ that has an inverse is called invertible and is determined by $f$, denoted by $f^{1}$ read $f$ inverse.
Inverse operations are the opposite of direct variation functions.
Let us find the inverse of $f(x)$ = $\frac{5x  8}{2x}$
Let $f(x)$ = $y$ = $\frac{5x  8}{2x}$
Interchange $x$ and $y$ in the above equation
$x$ = $\frac{5y  8}{2y}$
$2xy = 5y  8$
$2xy  5y$ = $ 8$
$y(2x  5)$ = $ 8$
$y$ = $\frac{8}{2x 5}$
Therefore, $f^{1}$ = $\frac{8}{2x  5}$
Function is defined as one quantity associated with another. By translation of a graph, we mean a shift in its location such that every point of the graph is moved the same distance and in the same direction. There are four ways to move the graph left, right, up and down and is explained below. Assuming that $c$ is a positive constant we have,
Equation 
Graph 
$y$ = $f(x) + c$  Translate c units upward 
$y$ = $f(x)  c$  Translate c units downward 
$y$ = $f(x + c)$  Translate c units to the left 
$y$ = $f(x  c)$  Translate c units to the right 