**Introduction** :

Quadratic functions are polynomials with the highest degree of 2. That is, it is a 2nd order polynomial. The general form of a quadratic function of x is y = ax2 + bx + c, for some numbers a, b
and c. The graphs of quadratic functions are called parabolas. The important characteristics of quadratic functions are found from their graphs.

Examples of some quadratic functions:

(a) y = x2 + 2x + 3 is a quadratic function of x.

(b) f(t) = t2 - t +10 is quadratic in t.

Quadratic Function to Parabola : Domain and Range of a Quadratic Function

The quadratic function can be written in the form y = a(x-p)2 + q.

We will look into the working of how this form is derived in a little while, but first let me tell you what this form means. This form is a common form of a parabola with it's vertex at the point
(p,q) and this parabola will open upwards or downwards depending upon the sign of coefficient 'a'. The parabola y = a(x-p)2 + q open upwards if a > 0 and downwards if a < 0. Look at the
figure changing with respect to the value of 'a' ( here the parabola has it's vertex at the origin (0,0))

Translations for Domain and Range of a Quadratic Function

Now, as I said, let's look into the working of how the form y = a(x-h)2 + k is derived from y = ax2 + bx + c

f(x) = ax2 + bx + c = a[x2 + x`b/a` + `c/a`] = a[(x + `b/(2a)`)2 - (`b/(2a)`)2 + `c/a`] = a[(x + `b/(2a)`)2 - (`(b^2 - 4ac)/(4a^2)`)]= a(x + `b/(2a)`)2 - (`(b^2 - 4ac)/(4a)`)

Here, h = - `b/(2a)` and k = - (`(b^2 - 4ac)/(4a)`)

This parabola has a vertex at ( - `b/(2a)`, -(`(b^2 - 4ac)/(4a)`))

So, the equation y = ax2 + bx + c becomes, `y =a(x -b/(2a))^2-(b^2-4ac)/(4a))` in the form y = a(x-h)2 + k

A translation of a geometric figure is a transformation in which every point is moved the same distance in the same direction. If the parabola y = ax2 is shifted h units to the right, and k units
upwards, the new function would be, y = a(x-h)2 + k. The figures below are self explanatory and it does not matter which translation is done first.

If parabola y = ax2 is shifted by h units to the right its new equation is y = a(x - h)^2 .

para-shift-x

Now if these translations are done together, then the final quadratic function would be, y = a(x-h)^2 + k.

If the parabola, y = ax^2 is shifted k units upwards, its new equation is y = ax^2 + k.

para-shift-y

Apparently these translations doesn't seem important regarding the domain and range of a Quadratic function, but as you can see, these translations changes the position of the parabola and hence
that of the quadratic function on the x-y plane. So, as we will see in the next section that the domain of the quadratic function is the whole x-axis, i.e., x in `(-oo,oo)` , this does not change
with any translation upwards or downwards, but the range of the function does vary. If the parabola is open upwards, and it is translated "k" units upwards, then the range of the function would
also exclude all the points from the present y-value to the translated y-value. That is, if the parabola, y = ax2 (range, y `in` `[0,oo)` ) is shifted "k" units upwards, then the new parabola
having the equation, y = ax2 + k would have it's range changed to y` in [k,oo)` excluding all the y-values from 0 to "k" excluding "k" i.e., y `!in [0,k)`

**Domain and Range of a Quadratic Function**

**Definition of the Domain of a Function**:

For a function f(x) defined by an expression with variable x, the implied domain of f(x) is the set of all real numbers variable x can take such that the expression defining the function is real.
The domain can also be given explicitly.

**Definition of the Range of a Function**:

The range of f(x) is the set of all values that the function takes when x takes values in the domain.

Now let’s take a look at the domain and range of a quadratic function:

From the calculations above

The equation y = ax2 + bx + c becomes, `y =a(x -b/(2a))^2-(b^2-4ac)/(4a))` in the form y = a(x-h)2 + k

Here, h = - `b/(2a)` and k = - (`(b^2 - 4ac)/(4a)`)

Now, for the domain of the function f(x), we can very well understand that any real value of x will give a real y. So, the domain of the function is x `in (-oo, oo)`

For the range of the function, let's examine the equation, y = (ax-h)2 + k

The term (x - h)^2 is either positive or zero. Hence (x - h)2 `>=` 0

CASE I : If a > 0, let's multiply both sides of the inequality above by 'a':

`=>` a(x - h)2 `>=` 0

Add k to both sides of the inequality

`=>` a(x - h)^2 + k `>=` k

The left side is the expression of the function, f(x) = a(x - h)2 + k , hence

`=>` f(x) `>=` k

The above result tells us that f(x) has a minimum value equal to k.

This also tells us that the range of f(x) is given by y `in` [ k , `oo` )

CASE II : If a < 0, let's multiply both sides of the inequality (x - h)^2 `>=` 0 by 'a' and therefore change the inequality symbol.

`=>` a(x - h)2 `<=` 0

Add k to both sides of the inequality

`=>` a(x - h)2 + k `<=` k

The left side is the formula of the function, f(x) = a(x - h)^2 + k , hence

`=>` f(x) `<=` k

The above result tells us that f(x) has a maximum value equal to k.

This also tells us that the range of f(x) is given by y `in` ( `-oo` , k]

Note here, that k = f(h).