The quadratic function has maximum power of
equal to
:
.
When equated to zero, the quadratic function becomes the quadratic equation:
, in which coefficient
is non-zero.
The solution of the quadratic equation is:

The above solution to the quadratic is well known to high-school algebra students. The proof of the solution is usually presented to students as completion of the square, not presented here. To me completion of the square seemed too complicated. To help you acquire a new respect for the quadratic, presented below are more ways to solve it.
Solving the quadratic[edit]
The depressed quadratic
The depressed cubic leads to a solution of the cubic and the depressed quartic leads to a solution of the quartic. This paragraph shows that the depressed quadratic leads to a solution of the quadratic.
To produce the depressed function, let

where G is the degree of the function. For the quadratic G = 2. Let

Substitute (3) into (1) and expand:





In the depressed quadratic above the coefficient of
is
and that of
is
.


Substitute (5) into (3) and the result is the solution in (2).
p +- q
Let one value of
be
and another value of
be
.
Substitute these values into
above and expand.



Substitute
into


By observation and elementary deduction
You have drawn a few curves by hand and suppose that each curve is symmetrical about the vertical line through a minimum point.
Given
you suppose that
.

You have found the stationary point without using calculus. Continue as per calculus below.
By calculus
The derivative of
is
which equals
at a stationary point.
At the stationary point
.
Prove that the function is symmetrical about the vertical line through
.
Let
Substitute (12) in (1) and expand:

Let
Substitute (13) in (1) and expand

The expansion is the same for both values of x, (12) and (13). The curve is symmetrical.
If the function equals 0, then

Substitute this value of p in (12) and the result is (2).
By movement of the vertex
Begin with the basic quadratic:
.
Move vertex from origin
to
.
.
.
.
.
This equation is in the form of the quadratic
where:
.
Therefore
and
is the X coordinate of the vertex in the new position.
Continue as per calculus above.
p + qi
Let
where
Substitute this value of
into (1) and expand:

Terms containing
From (14),
and
Terms without
From (15) and (16):

This method shows the imaginary value
coming into existence to help with intermediate calculations and then going away before the end result appears.
Defining the quadratic[edit]
The quadratic function is usually defined as the familiar polynomial (1). However, the quadratic may be defined in other ways. Some are shown below:
By three points
If three points on the curve are known, the familiar polynomial may be deduced. For example, if three points
are given and the three points satisfy
, the values
may be calculated.

The solution of the three equations
gives the equation
.
If the three points were to satisfy
, the equation would be
.
By two points and a slope
Given two points
and
and the slope at
, calculate
.

Slope = 2Ax + B, therefore

The solution of the three equations
gives the polynomial
.
By movement of the vertex
Begin with the basic quadratic
.
If
has the value
, then
and the height of
above the vertex =
.
If we move the vertex to
, then the equation becomes
.
If
has the value
, then
and the height of
above the vertex =
.
The curve
and the curve
have the same shape.
It's just that the vertex of the former
has been moved to the vertex of the latter
.
The latter equation expanded becomes
.
The example under "two points and a slope" above is
.
Therefore

The example
may be expressed as

For proof, expand:

By compliance with the standard equation of the conic section
The quadratic function can comply with the format:
(See The General Quadratic below.)
For example, the function
can be expressed as:
or:
To express a valid quadratic in this way, both
or both
must be non-zero.
By a point and a straight line
The point is called the focus and the line is called the directrix.
The distance from point to line is non-zero.
The quadratic is the locus of a point that is equidistant from both focus and line at all times.
When the quadratic is defined in this way, it is usually called a parabola.
Quadratic as Parabola[edit]
Let the
have coordinates
Let the
have equation:
Let the point
be equidistant from both focus and directrix.
Distance from
to focus
.
Distance from
to directrix
.
By definition these two lengths are equal.
Let this equation have the form:
Therefore:

Given
calculate
.
There are two equations with two unknowns

The solutions are:

Graph of quadratic function

showing :
* X and Y intercepts in red,
* vertex and axis of symmetry in blue,
* focus and directrix in pink.
If the quadratic equation is expressed as
then:
The focus is the point
, and
The directrix has equation:
.
The
is exactly half-way between focus and directrix.
Vertex is the point
.
distance from directrix to focus.
Distance from vertex to focus
.
If the curve has equation
, then the vertex is at the origin
.
If the focus is the point
, then
and the equation
becomes
.
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An example with vertical focus
Figure 3: Graph of quadratic function with vertical focus

showing :
* vertex at (4,-1),
* focus at (4,1),
* directrix at y = -3.
Let
Directrix has equation:
. Focus has coordinates
.
This example has equation:
or
or
. See Figure 3.
Distance from vertex to focus =
Or:
Vertex has coordinates
Distance from vertex to focus
.
Curve has shape of
with vertex moved to
Quadratic with horizontal focus
Let the
have coordinates
Let the
have equation:
Let the point
be equidistant from both focus and directrix.
Distance from
to focus
.
Distance from
to directrix
.
By definition these two lengths are equal.
Let this equation have the form:
Therefore:

Given
calculate
There are two equations with two unknowns

The solutions are:

If the quadratic equation is expressed as
then:
The focus is the point
, and
The directrix has equation:
.
The
is exactly half-way between focus and directrix.
Vertex is the point
.
distance from directrix to focus.
Distance from vertex to focus
.
If the curve has equation
, then the vertex is at the origin
.
If the focus is the point
, then
and the equation
becomes
.
An example with horizontal focus
Figure 4: Graph of quadratic function with horizontal focus

showing :
* vertex at (-1,4),
* focus at (1,4),
* directrix at x = -3.
Let
Directrix has equation:
. Focus has coordinates
.
This example has equation:
or
or
. See Figure 4.
Distance from vertex to focus =
Given equation
calculate
.
Method 1. By algebra
Put equation in form:
where

Method 2. By analytical geometry
Distance from vertex to focus
Put equation in
-
form:

Vertex is point
Focus is point
Directrix has equation:
The Parabola[edit]
Figure 1: The Parabola 
Focus at point

Vertex at origin

Directrix is line

By definition

and

In Figure 1

See Figure 1. For simplicity values are defined as shown. Let an arbitrary point on the curve be
.
By definition,
. This expression expanded gives:
and slope =
.
If the equation of the curve is expressed as:
, then
.
Let a straight line through the focus intersect the parabola in two points
and
.

where
is the slope of line DB in Figure 1.

Characteristics of the Parabola
The parabola is a grab-bag of many interesting facts.
We prove first that the tangents at
and
are perpendicular.

The product of
and
. Therefore, the tangents (lines AB and AD in Figure 1) are perpendicular.
Second, we prove that the two tangents intersect on the directrix. Using
and
:

The
coordinate of the point of intersection satisfies both
and
. Therefore,

is the mid-point between
and
. Point A in Figure 1 has coordinates
Check our work:

The tangents intersect at
. They intersect on the directrix where
.
See Tangents perpendicular and oscillating.
Third, we prove that the triangle defined by the three points
and
is a right triangle.
Slope of line
.
Slope of line
.

The product of
and
is
. Therefore the two sides
are perpendicular
and the triangle
in Figure 1 is a right triangle with the right angle at
.
Fourth, we prove that the two lines
are perpendicular.
Point
. Point
.
Using slope
Slope of line
.
Slope of line
Therefore the two lines
are perpendicular.
More about the Parabola[edit]
Figure 1: The Parabola 
Focus at point

Vertex at origin

Directrix is line

By definition

and

In Figure 1

In the last section we proved several points about the parabola, beginning with line
and moving towards point
on the directrix. In this section, we prove the reverse, beginning with point
and moving towards line
.
Let
be any point on the directrix
.
Using
and any line through
is defined as
where
is the slope of the line.
Let this line intersect the parabola
. (In Figure 1, p = 1.)

The above defines the
coordinate/s of any line through
that intersect/s the parabola. We are interested in the tangent, a line that intersects in only one place. Therefore,
the discriminant is 0.

where
Slope of tangent1 =
(In Figure 1, tangent1 is the line
.)
Slope of tangent2 =
(In Figure 1, tangent2 is the line
.)
Prove that tangent1 and tangent2 are perpendicular.

The product of the two slopes is -1. Therefore, the two tangents are perpendicular.
From (24), we chose a value of
that made the discriminant 0. Therefore

(In Figure 1,
is the slope of line
. This statement agrees with
proved in the last section.)
We have a line joining the two points
. Calculate the intercept on the
axis.
Using
,

The line joining the two points
passes through the focus
.
Figure 2: The Parabola 
Lines

are parallel.
Line

divides area

into two halves equal by area.
Two lines parallel
In Figure 2 tangents
and
intersect at point
on the directrix.
The line
has value
. The line
is tangent to the curve at
.
The slope of tangent
. The slope of line
also
.
Therefore two lines
are parallel.
Area DGBHFD
where
Line
. The integral of this value
.
Area under line
![{\displaystyle {\begin{aligned}x_{1}&\\=\ \ \ \ &[m{\frac {x^{2}}{2}}+px]=8Rmmpp+4Rpp\\x_{2}&\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31e03d8bd1a6c8d46687dfc1c96dcb0540ae85e3)
Area under curve
![{\displaystyle {\begin{aligned}x_{1}&\\=\ \ \ \ &[{\frac {x^{3}}{12p}}]={\frac {16Rmmpp+4Rpp}{3}}\\x_{2}&\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc905bf09fa48c89748b00685c4d138ad2b485c3)
Area

Similarly it can be shown that Area
Therefore line
splits area
into two halves equal by area.
Reflectivity of the Parabola[edit]
Figure 1: The Parabola 
Focus at point

Vertex at origin

Directrix is line

By definition

and

In Figure 1

See Figure 1.
is a right triangle and point
is the midpoint of line
.
is congruent with
, and
.
.
1. Any ray of light emanating from a point source at F touches the parabola at B and is reflected away from B on a line that is always perpendicular to the directrix.
is the angle of incidence and
is the angle of reflection.
2. The path from
through focus to vertex and back to focus has length
.
The path from
to
to
has length
all paths to and from focus have the same length.
Using the Quadratic[edit]
In Theory
1) The quadratic may be used to examine itself.
Let a quadratic equation be:
.
Let the equation of a line be:
.
Let the line intersect the quadratic at
.
Therefore:

Let the line intersect the curve in exactly one place. Therefore
must have exactly one value and the discriminant is
.

A line that touches the curve at
has slope
.
Therefore the slope of the curve at
is
. This examination of the curve has produced the slope of the curve without using calculus.
Consider the curve:
. The aim is to calculate the slope of the curve at an arbitrary point
.

If
is to have exactly one value, discriminant
.
Therefore
The slope of the curve at an arbitrary point
.
For more information see earlier version of "Using the Quadratic."
2) The quadratic may be used to examine other curves, for example, the circle.
Define a circle of radius 5 at the origin:
Move the circle to
We want to know the values of
that contain the circle, that is, the values of
for each of which there is only one value of
.
Put the equation of the circle into a quadratic in
.
There is exactly one value of
if the discriminant is
.
Therefore
These values of
make sense because we expect the values of
to be
.
This process has calculated a minimum point and a maximum point without calculus.
3) The formula remains valid for
and/or
equal to
. Under these conditions you probably won't need the formula. For example
can be factored by inspection as
.
4) The quadratic can be used to solve functions of higher order.
One of the solutions of the cubic depends on the solution of a sextic in the form
. This is the quadratic
where
.
The cubic function
produces the depressed function
.
The quadratic
is solved as
.
The roots of the depressed function are
.
Using
In this example the quadratic, as part of the depressed function, simplified the solution of the cubic.
The quartic function
produces the depressed function
which is the quadratic
where
.
5) The quadratic appears in Newton's Laws of Motion:
The General Quadratic[edit]
See Quadratic Equation:"Quadratic as Parabola" above.
See also Parabola:"Reverse-Engineering the Parabola", Method 2.
Reverse-Engineering the Parabola[edit]
See Parabola:"Reverse-Engineering the Parabola".
Area enclosed between parabola and chord[edit]
See Parabola:"Area enclosed between parabola and chord".
Gallery[edit]
Graph of quadratic function  showing basic features : * X and Y intercepts * vertex at (-2,-9), * axis of symmetry at x = -2.
|
Graph of quadratic function  showing : * X and Y intercepts in red, * vertex and axis of symmetry in blue, * focus and directrix in pink.
|