Quadratic Equation

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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 .


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


Area under curve


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.