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.
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).
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By observation and elementary deduction[edit | edit source]
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.
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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).
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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.
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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.
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The quadratic function is usually defined as the familiar polynomial (1). However, the quadratic may be defined in other ways. Some are shown below:
Figure 1: Diagram illustrating 2 quadratic curves that share 3 common 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
.
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Figure 1: Diagram illustrating quadratic curve defined by 2 points and slope at 1 point. Slope of curve at point  is 
Given two points and and the slope at , calculate .

Slope = 2Ax + B, therefore

The solution of the three equations gives the polynomial .
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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 .
Consider function .
Therefore

The example may be expressed as

For proof, expand:

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By compliance with the standard equation of the conic section[edit | edit source]
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.
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By a point and a straight line[edit | edit source]
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.
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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
.
An example with vertical focus[edit | edit source]
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
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Quadratic with horizontal focus[edit | edit source]
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[edit | edit source]
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

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

Vertex is point
Focus is point
Directrix has equation:
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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[edit | edit source]
The parabola is a grab-bag of many interesting facts.
Two tangents perpendicular[edit | edit source]
Figure 2: The Parabola  Directrix is line  Tangents  and  intersect at point  where they are perpendicular.
We prove first that the tangents at and are perpendicular.

The product of and . Therefore, the tangents (lines AB and AD in Figure 2) are perpendicular.
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Two tangents intersect on directrix[edit | edit source]
Figure 3: Diagram illustrating right triangle with right angle at focus 
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 .
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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
.
Area
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

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Area
Similarly it can be shown that Area
Therefore line splits area into two halves equal by area.
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Reflectivity of the Parabola[edit | edit source]
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:
See Quadratic Equation:"Quadratic as Parabola" above.
See also Parabola:"Reverse-Engineering the Parabola", Method 2.
Reverse-Engineering the Parabola[edit | edit source]
See Parabola:"Reverse-Engineering the Parabola".
Area enclosed between parabola and chord[edit | edit source]
See Parabola:"Area enclosed between parabola and chord".
Graph of quadratic function  showing basic features : * X and Y intercepts * vertex at (-2,-9), * axis of symmetry at x = -2.
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Graph of quadratic function  showing : * X and Y intercepts in red, * vertex and axis of symmetry in blue, * focus and directrix in pink.
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