Quadratic Equation

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 quadraticEdit



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 quadraticEdit


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 ParabolaEdit


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 ParabolaEdit

 
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 ParabolaEdit

 
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 ParabolaEdit


 
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 QuadraticEdit


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 QuadraticEdit


See Quadratic Equation:"Quadratic as Parabola" above.


See also Parabola:"Reverse-Engineering the Parabola", Method 2.

Reverse-Engineering the ParabolaEdit


See Parabola:"Reverse-Engineering the Parabola".

Area enclosed between parabola and chordEdit


See Parabola:"Area enclosed between parabola and chord".

GalleryEdit

 
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.