The quadratic function has maximum power of $x$ equal to $2$:
 $Ax^{2}+Bx+C$.
When equated to zero, the quadratic function becomes the quadratic equation:
 $Ax^{2}+Bx+C=0\ \dots \ (1)$, in which coefficient $A$ is nonzero.
The solution of the quadratic equation is:
 $x={\frac {B\pm {\sqrt {B^{2}4AC}}}{2A}}\ \dots \ (2)$
The above solution to the quadratic is well known to highschool 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
 $x={\frac {B+t}{GA}}$
where G is the degree of the function. For the quadratic G = 2. Let
 $x={\frac {B+t}{2A}}\ \dots \ (3)$
Substitute (3) into (1) and expand:
 $A({\frac {B+t}{2A}})({\frac {B+t}{2A}})+B({\frac {B+t}{2A}})+C\ \dots \ (4)$
 $(4)\ *\ 4A,\ 4AA({\frac {B+t}{2A}})({\frac {B+t}{2A}})+4AB({\frac {B+t}{2A}})+4AC$
 $(B+t)(B+t)+2B(B+t)+4AC$
 $B^{2}2Bt+t^{2}2B^{2}+2Bt+4AC$
 $t^{2}B^{2}+4AC$
In the depressed quadratic above the coefficient of $t^{2}$ is $1$ and that of $t$ is $0$.
 $t^{2}=B^{2}4AC$
 $t=\pm {\sqrt {B^{2}4AC}}\ \dots \ (5)$
Substitute (5) into (3) and the result is the solution in (2).

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 $f(x)=Ax^{2}+Bx+C$ you suppose that $f(p+q)=f(pq)$.
 ${\begin{aligned}A(p+q)(p+q)+B(p+q)+C=&\ A(pq)(pq)+B(pq)+C\\\\A(pp+2pq+qq)+Bp+Bq=&\ A(pp2pq+qq)+BpBq\\\\A2pq+Bq=&\ A2pqBq\\\\2A2pq+2Bq=&\ 0\\\\2q(2Ap+B)=&\ 0\\\\2Ap+B=&\ 0\\\\p=&\ {\frac {B}{2A}}\end{aligned}}$
You have found the stationary point without using calculus. Continue as per calculus below.

The derivative of $(1)$ is $2Ax+B$ which equals $0$ at a stationary point.
At the stationary point $x={\frac {B}{2A}}$.
Prove that the function is symmetrical about the vertical line through $x={\frac {B}{2A}}$.
Let $x={\frac {B}{2A}}+p\ \dots \ (12)$
Substitute (12) in (1) and expand:
 ${\frac {4AApp+4ACBB}{4A}}$
Let $x={\frac {B}{2A}}p\ \dots \ (13)$
Substitute (13) in (1) and expand
 ${\frac {4AApp+4ACBB}{4A}}$
The expansion is the same for both values of x, (12) and (13). The curve is symmetrical.
If the function equals 0, then
 ${\begin{aligned}+4AApp+4ACBB=&\ 0\\\\4AApp=&\ BB4AC\\\\pp=&\ {\frac {BB4AC}{4AA}}\\\\p=&\ {\frac {\pm {\sqrt {B^{2}4AC}}}{2A}}\\\end{aligned}}$
Substitute this value of p in (12) and the result is (2).

Begin with the basic quadratic: $y=Ax^{2}$.
Move vertex from origin $(0,0)$ to $(h,k)$.
$yk=A(xh)^{2}$.
$yk=A(x^{2}2hx+h^{2})$.
$yk=Ax^{2}2Ahx+Ah^{2}$.
$y=Ax^{2}2Ahx+Ah^{2}+k$.
This equation is in the form of the quadratic $y=Ax^{2}+Bx+C$ where:
$B=2Ah;\ C=Ah^{2}+k$.
Therefore $h={\frac {B}{2A}}$ and $h$ is the X coordinate of the vertex in the new position.
Continue as per calculus above.

Let $x=p+qi$ where $i={\sqrt {1}}$
Substitute this value of $x$ into (1) and expand:
 ${\begin{aligned}A(p+qi)(p+qi)+B(p+qi)+C\\\\A(p^{2}+2pqi+(qi)^{2})+Bp+Bqi+C\\\\Ap^{2}+2ApqiAq^{2}+Bp+Bqi+C\end{aligned}}$
Terms containing $i=2Apqi+Bqi\ \dots \ (14)$
From (14), $2Ap+B=0$ and $p={\frac {B}{2A}}\ \dots \ (15)$
Terms without $i=Ap^{2}Aq^{2}+Bp+C\ \dots \ (16)$
From (15) and (16):
 ${\begin{aligned}Aq^{2}=&\ Ap^{2}+Bp+C\\\\=&\ A({\frac {B}{2A}})({\frac {B}{2A}})+B({\frac {B}{2A}})+C\\\\q^{2}=&\ {\frac {B^{2}}{4AA}}{\frac {B^{2}}{2AA}}+{\frac {C}{A}}\\\\=&\ {\frac {B^{2}}{4AA}}{\frac {2B^{2}}{4AA}}+{\frac {4AC}{4AA}}\\\\=&\ {\frac {B^{2}+4AC}{4AA}}\\\\=&\ {\frac {1(B^{2}4AC)}{4AA}}\\\\q=&\ {\frac {\pm \ i{\sqrt {B^{2}4AC}}}{2A}}\\\\x=&\ (p)+(q)i\\\\=&\ ({\frac {B}{2A}})\pm ({\frac {i{\sqrt {B^{2}4AC}}}{2A}})i\\\\=&\ {\frac {B\pm {\sqrt {B^{2}4AC}}}{2A}}\end{aligned}}$
This method shows the imaginary value $i$ coming into existence to help with intermediate calculations and then going away before the end result appears.

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 $(5,40),\ (4,13),\ (7,76)$ are given and the three points satisfy $y=f(x)$, the values $A,B,C$ may be calculated.
 ${\begin{aligned}A(5)(5)+B(5)+C=&\ 40\\A(4)(4)+B(4)+C=&\ 13\\A(7)(7)+B(7)+C=&\ 76\\25A5B+C=&\ 40\ \dots \ (17)\\16A+4B+C=&\ 13\ \dots \ (18)\\49A+7B+C=&\ 76\ \dots \ (19)\\\end{aligned}}$
The solution of the three equations $(17),\ (18),\ (19)$ gives the equation $y=2x^{2}x15$.
If the three points were to satisfy $x=f(y)$, the equation would be
$x={\frac {2}{189}}y^{2}{\frac {169}{189}}y+{\frac {2615}{189}}$.

Figure 1: Diagram illustrating quadratic curve defined by 2 points and slope at 1 point. Slope of curve at point $(1,4.8)$ is $3.2.$
Given two points $(4,1.3)$ and $(1,4.8)$ and the slope at $(1,4.8)=3.2$, calculate $A,B,C$.
 ${\begin{aligned}A(4)(4)+B(4)+C=\ 1.3\\A(1)(1)+B(1)+C=\ 4.8\\\end{aligned}}$
Slope = 2Ax + B, therefore
 ${\begin{aligned}2A(1)+B=&\ 3.2\\\\16A4B+C=&\ 1.3\ \dots \ (20)\\A+B+C=&\ 4.8\ \dots \ (21)\\2A+B=&\ 3.2\ \dots \ (22)\\\end{aligned}}$
The solution of the three equations $(20),(21),(22)$ gives the polynomial $0.5x^{2}+2.2x+2.1\ \dots \ (23)$.

By compliance with the standard equation of the conic section[edit  edit source]
The quadratic function can comply with the format: $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.$
(See The General Quadratic below.)
For example, the function $y=3x^{2}+5x7$ can be expressed as:
$(3)x^{2}+(0)xy+(0)y^{2}+(5)x+(1)y+(7)=0$ or:
$3x^{2}+5x7y=0.$
To express a valid quadratic in this way, both $A,E$ or both $C,D$ must be nonzero.

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 nonzero.
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.

Let the $focus$ have coordinates $(p,q).$
Let the $directrix$ have equation: $y=k.$
Let the point $(x,y)$ be equidistant from both focus and directrix.
Distance from $(x,y)$ to focus $={\sqrt {(xp)^{2}+(yq)^{2}}}$.
Distance from $(x,y)$ to directrix $=yk$.
By definition these two lengths are equal.
${\sqrt {(xp)^{2}+(yq)^{2}}}=yk$
$x^{2}2px+p^{2}+y^{2}2qy+q^{2}=y^{2}2ky+k^{2}$
$(2q2k)y=x^{2}2px+p^{2}+q^{2}k^{2}$
$y={\frac {x^{2}}{2q2k}}+{\frac {2px}{2q2k}}+{\frac {p^{2}+q^{2}k^{2}}{2q2k}}$
Let this equation have the form: $y=Ax^{2}+Bx+C$
Therefore:
 ${\begin{aligned}A&={\frac {1}{2q2k}}\\B&={\frac {2p}{2q2k}}\\C&={\frac {p^{2}+q^{2}k^{2}}{2q2k}}\end{aligned}}$
Given $A,B,C$ calculate $p,q,k$.
$B={\frac {2p}{2q2k}}=2pA;\ p={\frac {B}{2A}}$
There are two equations with two unknowns $q,k:$
 ${\begin{aligned}A&(2q2k)1=0\\C&(2q2k)p^{2}q^{2}+k^{2}=0\end{aligned}}$
The solutions are:
 ${\begin{aligned}p&={\frac {B}{2A}}\\q&={\frac {1(B^{2}4AC)}{4A}}\\k&={\frac {1(B^{2}4AC)}{4A}}\end{aligned}}$
Graph of quadratic function
$y=x^{2}2x3$ 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 $y=Ax^{2}+Bx+C$ then:
The focus is the point $({\frac {B}{2A}},{\frac {1(B^{2}4AC)}{4A}})$, and
The directrix has equation: $y={\frac {1(B^{2}4AC)}{4A}}$.
The $vertex$ is exactly halfway between focus and directrix.
Vertex is the point $({\frac {B}{2A}},{\frac {(B^{2}4AC)}{4A}})$.
$A={\frac {1}{2q2k}};\ 2q2k={\frac {1}{A}};\ qk={\frac {1}{2A}}=$ distance from directrix to focus.
Distance from vertex to focus $={\frac {1}{4A}}$.
If the curve has equation $y=Ax^{2}$, then the vertex is at the origin $(0,0)$.
If the focus is the point $(0,q)$, then $q={\frac {1}{4A}};\ A={\frac {1}{4q}}$ and the equation $y=Ax^{2}$ becomes $y={\frac {x^{2}}{4q}}$.
An example with vertical focus[edit  edit source]
Figure 3: Graph of quadratic function with vertical focus $8(y+1)=(x4)^{2}$ showing : * vertex at (4,1), * focus at (4,1), * directrix at y = 3.
Let $(p,q)=(4,1),\ k=3.$
Directrix has equation: $y=3$. Focus has coordinates $(4,1)$.
${\begin{aligned}A&={\frac {1}{2q2k}}={\frac {1}{2(1)2(3)}}={\frac {1}{2+6}}={\frac {1}{8}}\\B&={\frac {2p}{2q2k}}={\frac {2(4)}{8}}=1\\C&={\frac {p^{2}+q^{2}k^{2}}{2q2k}}={\frac {16+19}{8}}=1\end{aligned}}$
This example has equation: $y={\frac {1}{8}}x^{2}x+1$ or $8y=x^{2}8x+8$ or $8(y+1)=(x4)^{2}$. See Figure 3.
Distance from vertex to focus = ${\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}={\frac {8}{4}}=2.$
Or:
Vertex has coordinates $(4,1).$
Distance from vertex to focus $=2={\frac {1}{4A}};\ A={\frac {1}{8}}$.
Curve has shape of $y={\frac {1}{8}}x^{2}$ with vertex moved to $(4,1).\ y(1)={\frac {1}{8}}(x4)^{2};\ 8(y+1)=(x4)^{2}.$

Quadratic with horizontal focus[edit  edit source]
Let the $focus$ have coordinates $(p,q).$
Let the $directrix$ have equation: $x=k.$
Let the point $(x,y)$ be equidistant from both focus and directrix.
Distance from $(x,y)$ to focus $={\sqrt {(xp)^{2}+(yq)^{2}}}$.
Distance from $(x,y)$ to directrix $=xk$.
By definition these two lengths are equal.
${\sqrt {(xp)^{2}+(yq)^{2}}}=xk$
$x^{2}2px+p^{2}+y^{2}2qy+q^{2}=x^{2}2kx+k^{2}$
$(2p2k)x=y^{2}2qy+p^{2}+q^{2}k^{2}$
$x={\frac {y^{2}}{2p2k}}+{\frac {2qy}{2p2k}}+{\frac {p^{2}+q^{2}k^{2}}{2p2k}}$
Let this equation have the form: $x=Ay^{2}+By+C$
Therefore:
 ${\begin{aligned}A&={\frac {1}{2p2k}}\\B&={\frac {2q}{2p2k}}\\C&={\frac {p^{2}+q^{2}k^{2}}{2p2k}}\end{aligned}}$
Given $A,B,C$ calculate $p,q,k.$
$B={\frac {2q}{2p2k}}=2qA;\ q={\frac {B}{2A}}$
There are two equations with two unknowns $p,k:$
 ${\begin{aligned}A&(2p2k)1=0\\C&(2p2k)p^{2}q^{2}+k^{2}=0\end{aligned}}$
The solutions are:
 ${\begin{aligned}p&={\frac {1(B^{2}4AC)}{4A}}\\q&={\frac {B}{2A}}\\k&={\frac {1(B^{2}4AC)}{4A}}\end{aligned}}$
If the quadratic equation is expressed as $x=Ay^{2}+By+C$ then:
The focus is the point $({\frac {1(B^{2}4AC)}{4A}},{\frac {B}{2A}})$, and
The directrix has equation: $x={\frac {1(B^{2}4AC)}{4A}}$.
The $vertex$ is exactly halfway between focus and directrix.
Vertex is the point $({\frac {(B^{2}4AC)}{4A}},{\frac {B}{2A}})$.
$A={\frac {1}{2p2k}};\ 2p2k={\frac {1}{A}};\ pk={\frac {1}{2A}}=$ distance from directrix to focus.
Distance from vertex to focus $={\frac {1}{4A}}$.
If the curve has equation $x=Ay^{2}$, then the vertex is at the origin $(0,0)$.
If the focus is the point $(p,0)$, then $p={\frac {1}{4A}};\ A={\frac {1}{4p}}$ and the equation $x=Ay^{2}$ becomes $x={\frac {y^{2}}{4p}}$.
An example with horizontal focus[edit  edit source]
Figure 4: Graph of quadratic function with horizontal focus $8(x+1)=(y4)^{2}$ showing : * vertex at (1,4), * focus at (1,4), * directrix at x = 3.
Let $(p,q)=(1,4),\ k=3.$
Directrix has equation: $x=3$. Focus has coordinates $(1,4)$.
${\begin{aligned}A&={\frac {1}{2p2k}}={\frac {1}{2(1)2(3)}}={\frac {1}{8}}\\B&={\frac {2q}{2p2k}}={\frac {2(4)}{8}}=1\\C&={\frac {p^{2}+q^{2}k^{2}}{2p2k}}={\frac {1+169}{8}}=1\end{aligned}}$
This example has equation: $x={\frac {1}{8}}y^{2}y+1$ or $8x=y^{2}8y+8$ or $8(x+1)=(y4)^{2}$. See Figure 4.
Distance from vertex to focus = ${\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}={\frac {8}{4}}=2.$
Given equation $8x=y^{2}8y+8$ calculate $p,q,k$.
Method 1. By algebra
Put equation in form: $x={\frac {1}{8}}y^{2}y+1$ where $A={\frac {1}{8}},\ B=1,\ C=1.$
 ${\begin{aligned}p&={\frac {1(B^{2}4AC)}{4A}}={\frac {1(1{\frac {1}{2}})}{4({\frac {1}{8}})}}={\frac {1{\frac {1}{2}}}{\frac {1}{2}}}=1\\q&={\frac {B}{2A}}={\frac {(1)}{2({\frac {1}{8}})}}={\frac {8}{2}}=4\\k&={\frac {1(B^{2}4AC)}{4A}}={\frac {1(1{\frac {1}{2}})}{\frac {1}{2}}}={\frac {1{\frac {1}{2}}}{\frac {1}{2}}}=3\end{aligned}}$

Method 2. By analytical geometry
Distance from vertex to focus $={\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}=2.$
Put equation in $semi$$reduced$ form:
 ${\begin{aligned}8x=y^{2}8y+8\\(y4)^{2}=y^{2}8y+16\\8x=(y4)^{2}8\\8x+8=(y4)^{2}\\8(x+1)=(y4)^{2}\end{aligned}}$
Vertex is point $(1,4).$
Focus is point $(1+2,4)=(1,4)=(p,q).$
Directrix has equation: $x=12=3=k.$




Figure 1: The Parabola $y={\frac {x^{2}}{4p}}$ Focus at point
$F\ (0,p)$ Vertex at origin
$(0,0)$ Directrix is line
$y=p$ By definition
$BF=BH$ and
$DF=DG$ In Figure 1
$p=1$
See Figure 1. For simplicity values are defined as shown. Let an arbitrary point on the curve be $(x,y)$.
By definition, ${\sqrt {(x0)^{2}+(yp)^{2}}}=y+p$. This expression expanded gives:
 $x^{2}4py=0;\ y={\frac {x^{2}}{4p}}$ and slope = ${\frac {x}{2p}}$.
If the equation of the curve is expressed as: $y=Kx^{2}$, then $K={\frac {1}{4p}};\ 4Kp=1;\ p={\frac {1}{4K}}$.
Let a straight line through the focus intersect the parabola in two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$.
 ${\begin{aligned}y=&\ {\frac {x^{2}}{4p}}=mx+p\\\\x^{2}=&\ 4pmx+4pp\\\\x^{2}4pmx4pp=&\ 0\\\\x=&\ {\frac {4pm\pm {\sqrt {16ppmm+16pp}}}{2}}\\\\=&\ {\frac {4pm\pm 4p{\sqrt {mm+1}}}{2}}\\\\=&\ 2pm\pm 2p{\sqrt {mm+1}}\\\\=&\ 2pm\pm 2pR\\\end{aligned}}$
where
 $m$ is the slope of line DB in Figure 1.
 ${\begin{aligned}R=&\ {\sqrt {m^{2}+1}}\\\\x_{1}=&\ 2pm+2pR\\\\x_{2}=&\ 2pm2pR\\\\y_{1}=&\ {\frac {x_{1}^{2}}{4p}}=2Rmp+2mmp+p\\\\y_{2}=&\ 2Rmp+2mmp+p\end{aligned}}$
Characteristics of the Parabola[edit  edit source]
The parabola is a grabbag of many interesting facts.
Two tangents perpendicular[edit  edit source]
Figure 2: The Parabola $y={\frac {x^{2}}{4}}$ Directrix is line $y=1$ Tangents $EDA$ and $CBA$ intersect at point $A$ where they are perpendicular.
We prove first that the tangents at $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are perpendicular.
 ${\begin{aligned}s=&\ {\frac {x}{2p}}\\\\s1=&\ {\frac {x_{1}}{2p}}=m+R\\\\s2=&\ mR\end{aligned}}$
The product of $s_{1}$ and $s_{2}=(m+R)(mR)=m^{2}R^{2}=m^{2}(m^{2}+1)=1$. Therefore, the tangents (lines AB and AD in Figure 2) are perpendicular.

Two tangents intersect on directrix[edit  edit source]
Figure 3: Diagram illustrating right triangle $GFH$ with right angle at focus $F.$
Third, we prove that the triangle defined by the three points $H\ (x1,p),G\ (x2,p)$ and $F\ (0,p)$ is a right triangle.
Slope of line $(0,p)...(x_{1},p)=s_{3}={\frac {p(p)}{0x_{1}}}={\frac {2p}{(2pm+2pR)}}={\frac {1}{m+R}}$.
Slope of line $(0,p)...(x_{2},p)=s_{4}={\frac {p(p)}{0x_{2}}}={\frac {2p}{(2pm2pR)}}={\frac {1}{mR}}$.
 $(s_{3})*(s_{4})={\frac {1}{m+R}}\ *\ {\frac {1}{mR}}={\frac {1}{m^{2}R^{2}}}={\frac {1}{1}}=1$
The product of $s_{3}$ and $s_{4}$ is $1$. Therefore the two sides $HF,GF$ are perpendicular
and the triangle $HFG$ in Figure 1 is a right triangle with the right angle at $F$.

Figure 1: The Parabola $y={\frac {x^{2}}{4p}}$ Focus at point
$F\ (0,p)$ Vertex at origin
$(0,0)$ Directrix is line
$y=p$ By definition
$BF=BH$ and
$DF=DG$ In Figure 1
$p=1$
In the last section we proved several points about the parabola, beginning with line $DFB$ and moving towards point $A$ on the directrix. In this section, we prove the reverse, beginning with point $A$ and moving towards line $DFB$.
Let $(k,p)$ be any point on the directrix $y=p$.
Using $y=mx+c,p=mk+c,\ c=pmk$ and any line through $(k,p)$
is defined as $y=sxpsk$ where $s$ is the slope of the line.
Let this line intersect the parabola $y={\frac {x^{2}}{4p}}$. (In Figure 1, p = 1.)
 ${\begin{aligned}y&\ ={\frac {x^{2}}{4p}}=sxpsk\\\\x^{2}&\ =4psx4pp4psk\\\\x^{2}&\ 4psx+4pp+4psk=0\ \dots \ (24)\\\\x^{2}&\ +(4ps)x+(4pp+4psk)=0\end{aligned}}$
The above defines the $X$ coordinate/s of any line through $(k,p)$ that intersect/s the parabola. We are interested in the tangent, a line that intersects in only one place. Therefore,
the discriminant is 0.
 ${\begin{aligned}16ppss4(4pp+4psk)=0\\\\16ppss16(pp+psk)=0\\\\ppsspkspp=0\\\\pssksp=0\\\\s=\ {\frac {k\pm {\sqrt {kk+4pp}}}{2p}}=\ {\frac {k\pm R}{2p}}\end{aligned}}$
where $R={\sqrt {kk+4pp}}$
Slope of tangent1 = $s_{1}={\frac {k+R}{2p}}$ (In Figure 1, tangent1 is the line $ABC$.)
Slope of tangent2 = $s_{2}={\frac {kR}{2p}}$ (In Figure 1, tangent2 is the line $ADE$.)
Prove that tangent1 and tangent2 are perpendicular.
 $s_{1}*s_{2}={\frac {k+R}{2p}}*{\frac {kR}{2p}}={\frac {kkRR}{4pp}}={\frac {kk(kk+4pp)}{4pp}}={\frac {4pp}{4pp}}=1$
The product of the two slopes is 1. Therefore, the two tangents are perpendicular.
From (24), we chose a value of $s$ that made the discriminant 0. Therefore
 ${\begin{aligned}x=&\ {\frac {B}{2A}}={\frac {4ps}{2}}=2ps\\\\x_{1}=&\ 2ps_{1}=2p*{\frac {k+R}{2p}}=k+R\\\\x_{2}=&\ kR\\\\y=&\ {\frac {x^{2}}{4p}}\\\\y_{1}=&\ {\frac {(k+R)^{2}}{4p}}={\frac {kk+2kR+RR}{4p}}={\frac {kk+2kR+kk+4pp}{4p}}={\frac {2kk+2kR+4pp}{4p}}={\frac {kk+kR+2pp}{2p}}\\\\y_{2}=&\ {\frac {kkkR+2pp}{2p}}\\\\m=&\ {\frac {y_{1}y_{2}}{x_{1}x_{2}}}={\frac {kR/p}{2R}}={\frac {k}{2p}}\end{aligned}}$
(In Figure 1, $m$ is the slope of line $DFB$. This statement agrees with $k=2mp$ proved in the last section.)
We have a line joining the two points $(x1,y1),(x_{2},y_{2})$. Calculate the intercept on the $Y$ axis.
Using $y=mx+c$,
 ${\begin{aligned}c=&\ ymx=y_{1}mx_{1}\\\\=&\ {\frac {kk+kR+2pp}{2p}}{\frac {k}{2p}}*(k+R)\\\\=&\ {\frac {kk+kR+2pp(kk+kR)}{2p}}\\\\=&\ {\frac {2pp}{2p}}=p\end{aligned}}$
The line joining the two points $(x_{1},y_{1}),\ (x_{2},y_{2})$ passes through the focus $(0,p)$.
Area $DGBHFD$
$x_{1}=2pm+2pR$
$x_{2}=2pm2pR$
where $R={\sqrt {m^{2}+1}}$
Line $DFB=mx+p$. The integral of this value $=m{\frac {x^{2}}{2}}+px$.
Area under line $DFB$
 ${\begin{aligned}x_{1}&\\=\ \ \ \ &[m{\frac {x^{2}}{2}}+px]=8Rmmpp+4Rpp\\x_{2}&\end{aligned}}$
Area under curve $DGB$
 ${\begin{aligned}x_{1}&\\=\ \ \ \ &[{\frac {x^{3}}{12p}}]={\frac {16Rmmpp+4Rpp}{3}}\\x_{2}&\end{aligned}}$
Area $DGBHFD$
 ${\begin{aligned}=&\ 8Rmmpp+4Rpp{\frac {16Rmmpp+4Rpp}{3}}\\\\=&\ {\frac {24Rmmpp+12Rpp16Rmmpp4Rpp}{3}}\\\\=&\ {\frac {8Rmmpp+8Rpp}{3}}\\\\=&\ {\frac {8Rp^{2}(m^{2}+1)}{3}}\end{aligned}}$

Area $DGHD$
Similarly it can be shown that Area $DGHD={\frac {4Rp^{2}(m^{2}+1)}{3}}$
Therefore line $AGH$ splits area $DGBHFD$ into two halves equal by area.


Reflectivity of the Parabola[edit  edit source]
In Theory
1) The quadratic may be used to examine itself.
Let a quadratic equation be: $y=x^{2}$.
Let the equation of a line be: $y=mx+c$.
Let the line intersect the quadratic at $(x_{1},y_{1})$.
Therefore:
 ${\begin{aligned}y_{1}=mx_{1}+c\\c=y_{1}mx_{1}\\y=x^{2}=mx+c=mx+(y_{1}mx_{1})\\x^{2}mxy_{1}+mx_{1}=0\end{aligned}}$
Let the line intersect the curve in exactly one place. Therefore $x$ must have exactly one value and the discriminant is $0$.
 ${\begin{aligned}m^{2}4(mx_{1}y_{1})=0\\m^{2}4mx_{1}+4y_{1}=0\\m^{2}4mx_{1}+4x_{1}^{2}=0\\(m2x_{1})^{2}=0\\m=2x_{1}\end{aligned}}$
A line that touches the curve at $x_{1},y_{1}$ has slope $2x_{1}$.
Therefore the slope of the curve at $x_{1},y_{1}$ is $2x_{1}$. This examination of the curve has produced the slope of the curve without using calculus.
Consider the curve: $y=Ax^{2}+Bx+C$. The aim is to calculate the slope of the curve at an arbitrary point $(x_{1},y_{1})$.
 ${\begin{aligned}Ax^{2}+Bx+C=mx+c\\Ax^{2}+Bx+C=mx+(y_{1}mx_{1})\\Ax^{2}+Bx+Cmxy_{1}+mx_{1}=0\\Ax^{2}+Bxmx+Cy_{1}+mx_{1}=0\\Ax^{2}+(Bm)x+(Cy_{1}+mx_{1})=0\end{aligned}}$
If $x$ is to have exactly one value, discriminant $=(Bm)^{2}4(A)(Cy_{1}+mx_{1})=0$.
Therefore
$BB2Bm+mm4AC+4Ay_{1}4Amx_{1}=0$
$mm4Ax_{1}m2Bm+BB4AC+4Ay_{1}=0$
$mm(4Ax_{1}+2B)m+(BB4AC+4Ay_{1})=0$
$m={\frac {(4Ax_{1}+2B)\pm {\sqrt {(4Ax_{1}+2B)^{2}4(BB4AC+4Ay_{1})}}}{2}}$
$m={\frac {(4Ax_{1}+2B)\pm {\sqrt {16AAx_{1}x_{1}+16ABx_{1}+4BB4BB+16AC16Ay_{1}}}}{2}}$
$m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {AAx_{1}x_{1}+ABx_{1}+ACAy_{1}}}}{2}}$
$m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {A(Ax_{1}^{2}+Bx_{1}+Cy_{1})}}}{2}}$
$m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {A(0)}}}{2}}$
$m=2Ax_{1}+B.$
The slope of the curve at an arbitrary point $(x_{1},y_{1})=2Ax_{1}+B$.
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:
${\sqrt {x^{2}+y^{2}}}=5$
$x^{2}+y^{2}=25$
Move the circle to $(8,3)$
$(x8)^{2}+(y3)^{2}=25$
$x^{2}16x+64+y^{2}6y+9=25$
$x^{2}16x+y^{2}6y+48=0$
We want to know the values of $x$ that contain the circle, that is, the values of $x$ for each of which there is only one value of $y$.
Put the equation of the circle into a quadratic in $y$.
$y^{2}6y+x^{2}16x+48=0$
$A=1,\ B=6,\ C=x^{2}16x+48$
There is exactly one value of $y$ if the discriminant is $0$.
Therefore
$(6)(6)4(x^{2}16x+48)=0$
$364(x^{2}16x+48)=0$
$9+x^{2}16x+48=0$
$x^{2}16x+39=0$
$(x3)(x13)=0$
$x_{1}=3,\ x_{2}=13$
These values of $x$ make sense because we expect the values of $x$ to be $8\pm 5$.
This process has calculated a minimum point and a maximum point without calculus.
3) The formula remains valid for $B$ and/or $C$ equal to $0$. Under these conditions you probably won't need the formula. For example $Ax^{2}+Bx$ can be factored by inspection as $x(Ax+b)$.
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 $Ax^{6}+Bx^{3}+C=0$. This is the quadratic
$AX^{2}+BX+C$ where $X=x^{3}$.
The cubic function $x^{3}+6x^{2}+13x+10$ produces the depressed function $t^{3}+9t=t(t^{2}+9)$.
The quadratic $t^{2}+9=0$ is solved as $t^{2}=9,\ t={\sqrt {9}}=\pm 3i$.
The roots of the depressed function are $t_{1}=0,\ t_{2}=3i,\ t_{3}=3i$.
Using $x={\frac {B+t}{3A}}$
$x_{1}={\frac {6+0}{3}}=2$
$x_{2}={\frac {6+3i}{3}}=2+i$
$x_{3}=2i$
In this example the quadratic, as part of the depressed function, simplified the solution of the cubic.
The quartic function $x^{4}+4x^{3}+17x^{2}+26x+11$ produces the depressed function
$t^{4}+176t^{2}256$ which is the quadratic $T^{2}+176T256$ where $T=t^{2}$.
5) The quadratic appears in Newton's Laws of Motion: $s=ut+{\frac {1}{2}}at^{2}$
See Quadratic Equation:"Quadratic as Parabola" above.
See also Parabola:"ReverseEngineering the Parabola", Method 2.
ReverseEngineering the Parabola[edit  edit source]
See Parabola:"ReverseEngineering the Parabola".
Area enclosed between parabola and chord[edit  edit source]
See Parabola:"Area enclosed between parabola and chord".
Graph of quadratic function $y=x^{2}+4x5$ showing basic features : * X and Y intercepts * vertex at (2,9), * axis of symmetry at x = 2.

Graph of quadratic function $y=x^{2}2x3$ showing : * X and Y intercepts in red, * vertex and axis of symmetry in blue, * focus and directrix in pink.
