A quadratic function is a second degree polynomial function. Quadratic functions are used to model many real life situations. It is widely used in Physics to describe motion of freely falling objects and paths traced by projectiles like Rockets.

Quadratic models also describe the shape of supporting cables in Suspension bridges. Graphically a quadratic function is represented by a parabola with interesting characteristics.

A second degree polynomial function of the form f(x) = ax2 + bx + c, where a, b and c are arbitrary constants and a $\neq$ 0 is defined as a quadratic function. Here ax2 is a quadratic term, bx a linear term and c a constant term. When a = 0, the function ceases to be quadratic. The other two constants b and c can be zero to be qualified as a quadratic function.

## Quadratic Function in Standard Form

The standard form of a quadratic function f(x) is f(x) = ax2 + bx + c where the terms are written in ascending order of powers.

Example:

p(x) = 2x2 + 3x - 5 All the three terms quadratic, linear and constant are present.

g(x) = $\frac{1}{2}$x2 Only the quadratic term is present.

h(x) = 3x2 - 5x Quadratic without a constant.

The standard from of a quadratic equation is ax2 + bx + c = 0, where the quadratic polynomial written in standard form is equated to zero.

Example:

x2 - 4x - 12 = 0

If a quadratic equation is not given in standard form it can be brought to the form using properties of equality like addition and subtraction.

1. The graph of a Quadratic function is represented by a Parabola is symmetric about a vertical line, which is called the Axis of symmetry, or simply the axis.

2. The Point where the graph of the function cuts the axis is called the vertex of the Parabola.

3. The parabola opens up, if the leading coefficient of the quadratic function f(x) = ax2 + bx + c is positive. In this case the vertex is the lowest point on the graph and hence the minimum function value occurs at the vertex.

4. The parabola opens down when the leading coefficient is negative. In this case the vertex is highest point on the graph and hence the maximum function value occurs at the vertex.

 f(x) = ax2 + bx + c a > 0 f(x) = ax2 + bx + c a < 0

5. The x intercepts of the graph represent the zeros of the quadratic function or the solution when it is expressed as an equation.
If the graph cuts the x-axis at two distinct points, the quadratic equation has two real solutions, one real solution if the graph touches the x-axis in one point and no solution if the graph does not intersect the x -axis.

6. The Domain of a Quadratic function is all real numbers. The range is [ Minimum, $\infty$) when the leading term is positive and ( -$\infty$, Maximum] when the leading term is negative.

The x - coordinate of the vertex of the graph of the quadratic function f(x) = ax2 + bx + c is given by the formula, x = $\frac{-b}{2a}$

The y coordinate can then be found by substituting the value found for x in the function rule.

Thus the formula for vertex as an ordered pair can be written as ($\frac{-b}{2a}$, $f($$\frac{-b}{2a}$$)$).

The axis of symmetry of the Quadratic function is given by x = $\frac{-b}{2a}$

When the quadratic function is defined in vertex form as f(x) = a(x - h)2 + k, the vertex of the function is given by the ordered pair (h, k) and the axis of symmetry is given by x = h. A quadratic function given in standard form can be rewritten into vertex form by completing the square.

The popular quadratic formula to be used for finding the roots of the equation ax2 + bx + c = 0 is
x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$.
This formula can also be used to determine the zeros of the quadratic function f(x) = ax2 + bx +c.

The x - coordinate of the vertex is the average of zeros of the quadratic function.

1. Find the vertex and axis of symmetry.
2. Find the x and y intercepts of the Quadratic Function.

The vertex and axis of symmetry can either be found using the formula x = $\frac{-b}{2a}$ or rewriting the function in vertex form by completing the square.
The x intercept can be found by solving the equation f(x) = 0, using the methods used to solve the quadratic equations.
The y - intercept is given by f(0) that is the function value at x = 0.

### Solved Example

Question: Find the vertex, axis of symmetry, the x and y intercepts and maximum or minimum and Domain and range of the quadratic function
f(x) = -x2 + 12x - 28
Solution:

Let us find the vertex using formula.

x = $\frac{-b}{2a}$ = $\frac{-12}{2(-1)}$ = 6

f(6) = -(6)2 + 12(6) - 28 = -36 + 72 - 28 = 8

Hence the vertex of the graph (6, 8)

Axis of symmetry is x = 6

y intercept f(0) = - 28

To find the x intercept, f(x) = -x2 + 12x - 28 = 0
x2 - 12x + 28 = 0
Using the quadratic formula x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

= $\frac{12\pm \sqrt{(-12)^{2}-4(1)(28))}}{2}$

= $\frac{12\pm \sqrt{144-112}}{2}$

Thus the two x intercepts are = 6 + √8  and 6 - √8
Since the leading term is negative, the graph is U down and the maximum value of the function is reached at the vertex.
Thus the maximum value of the function = 8.
The Domain of the function is all real numbers.
Range of the Quadratic function ( -$\infty$, 8]

## Solving Quadratic Functions by Graphing

The characteristics of a Quadratic function like Vertex, maximum or minimum value and intercepts can be estimated using the Graph of the function done on a Graphing calculator.

To graph a Quadratic function manually, the vertex and axis of symmetry are found using algebraic methods.

By doing the Graph it can be shown how the parabola is turned, up or down. The graph also verifies the number of x intercepts and confirms the situation of a quadratic function not having any real zeros.
 The graph of the example functiony = -x2 +12x -28 solved earlier is shown on the right. The Vertex and the x intercepts are marked on the graph. The shape of the graph and the estimated value of x intercepts agree with what wasdetermined algebraically.

Let us solve an application problem on Quadratic functions.

### Solved Example

Question: The height y (in feet) of a kicked ball is given by y = -$\frac{1}{135}$x2 + $\frac{9}{5}$x + 1.2 where x is the horizontal distance of the ball from the point where it was kicked.
(a) Find the height of the ball when it is kicked.
(b) What is the maximum height reached by the ball?
(c) Find the horizontal distance moved by the ball before it fell on the ground.

Solution:

(a)  The initial height is given by the y intercept of the quadratic function.  This when x = 0, y = 1.2
The ball was initially kicked at a height of 1.2 ft above the ground.
(b)   The maximum height is reached is given by the y - coordinate of the vertex.

x coordinate of the vertex = $\frac{-b}{2a}$ = $\frac{-\frac{9}{5}}{2(-\frac{1}{135})}$ = 121.5

y coordinate of the vertex = f(121.5) = 110.55 ft
Maximum height reached = 110.55 ft.

(c) The x - intercepts of the function found using quadratic formula are
x = -0.66 ft and 243.66 ft.
The ball moved a horizontal distance of 243.66 ft before it fell on the ground.