BY
ILONA MILEWSKA
&
ANNA UCHWAT
 “Archimedes is considered one
of the three greatest
mathematicians of all time
along with Newton and Gauss.
In his own time, he was known
as “the wise one,” “ the
master“, and “ the great
geometer” and his works and
inventions brought him fame
that lasts to his very day. He
was one of the last great
Greek mathematicians.” (Paul
Golba)
 In the Second Punic War the Romans were fighting against the people
of Syracuse. They had difficulty winning the bottle, because
Archimedes used a mathematical principle to burn the Roman ships.
He burned up the whole Roman fleet by discovering that on a curved
mirror he can concentrate the sun’s rays onto anything burnable with
enough intensity to cause the object to burst into flames in seconds.
 Archimedes would not burn the
enemies’ ships if the mirror that
he used was a plane mirror.
Sun’s rays would not reflect the
way that they would cross each
other in one point. This point
where all the sun rays are
crossing is called focus point.
Part 2
1. A
f ( x )   ( x  4) 2
2
To find the vertex using the standard form of the quadratic function
2
f ( x )  a ( x  h )  k , a  0 . There is a very easy rule to find it: Vertex = (h, k).
So, in the given quadratic function f ( x )   ( x  4 ) 2  2 the vertex is V =   4 ,  2  .
To find if the vertex   4 ,  2  is a maximum or minimum point of the quadratic
function without graphing, we have to determine a in the function. If a  0 , the vertex of
the parabola would be a minimum point, but if the a  0 , then the vertex of the parabola
would be a maximum point. In our function f ( x )   ( x  4 )  2 a is equal to -1 so it is
less than 0. We can conclude that the vertex   4 ,  2  is the maximum point of the
function.
2
1.B
f ( x)  x 2 x  2
2
To find the vertex of the function in the form
b
  b 
, f
 
2
a
2
a



the formula 
f ( x )  ax  bx  c , a  0
2
, we have to use
. First, we have to determine what is a, b and c, so a  1 , b   2
and c  2 . Now, the founded numbers we can already plug into the vertex
coordinates formula.
f ( x )  ax  2 x  2
2
f ( x)  x 2 x  2
2
x 
b

 (2)
2a
2 (1)

2
1
2
f ( x )  (1)  2 (1)  2  1
2
The vertex of the quadratic function is (1, 1).
To determine if the vertex of given function is minimum or maximum point of the
parabola of the quadratic function f ( x )  x  2 x  2 we need to find out if a  0 or a 
a  0 , then the vertex is a minimum point of the parabola, but when a  0 as in our
problem where a  1 then the vertex is the maximum point of the parabola of
quadratic function.
2
0
. If
2.A
f ( x )  ( x  2) 1
2
To graph the quadratic function we have to follow these steps:
STEP 1 Find if the graph curves up or down.
Because
, so 1>0,then we know that it is going to curve up.
STEP 2 Find the vertex.
As we know already from the standard form of the quadratic function we identify
vertex f ( x )  a ( x  h ) 2  k . In our function f ( x )  ( x  2 ) 2  1 , h   2 and k  1 , so the
coordinates of the vertex are   2 ,1  .
a 1
STEP 3 Find x and y intercepts.
Y-intercept
To find y intercept we have to replace x with 0 in the given function, because it is a point
where graph crosses the x-axis.
f ( x )  ( x  2) 1
2
f ( x )  ( 0  2 )  1  5 The y intercept is (0, 5)
2
X-intercept
To find x intercept we have to replace y with 0, because it is a point where graph crosses
the x-axis.
f ( x )  ( x  2) 1
2
0  ( x  2) 1
2
0  x 4 x  4  1
2
0  x 4 x  5
2
Now we have to solve this quadratic equation by using this formula
In our example a  1 , b  4 , and c  5 , so
x 
b
 b  b  4 ac
2
x
2a
b  4 ac
2
2a
x 
x 
4
( 4 )  4 1 5 
2
2 1 
4
4
2
Because under the square root is a negative number it means there is no real number
solution, no x-intercepts.
STEP 4 Graph the parabola
AXIS OF SYMMETRY
To find the axis of symmetry we have to take the x coordinate of the vertex or read it
from the graph. For this function b   2
2. B
f ( x)   x 1
2
To graph the function we have to follow these steps:
STEP 1 Find if the graph curve up or down.
Because a = -1, then a < 0 so we know it is going to be curve down.
STEP 2 Find the Vertex
As we already know that the function is in the form
the formula
b
  b 

, f
 
2
a
2
a



f ( x )  ax  bx  c
2
.
f ( x )  ax  bx  c
2
f ( x)   x 1
2
a   1, b  0 , c  1
x
b
2a

 (0)
2 (  1)

0
2
f ( x )   (0)  1  1
2
0
so the coordinates of the vertex are (0, 1)
so we have to use
STEP 3 Find x and y intercepts.
Y-intercept
To find y intercept we have to replace x with 0 in the given function, because it is
a point where graph crosses the x-axis
f ( x)   x 1
2
f ( x )   (0) 1  1
2
The y intercept is (0, 1)
X-intercept
To find x intercept we have to replace y with 0, because it is a point where graph
crosses the x-axis.
f ( x)   x 1
2
0   x 1
2
x 1  0
2
( x  1)( x  1)  0
x   1, x  1
The x intercepts are (-1, 0) and (1, 0).
STEP 4 Graph the parabola.
AXIS OF SYMMETRY
To find the axis of symmetry we have to take the x coordinate of the vertex or
read it from the graph. In this function x=0.
PART 3-1
The plane curve that emerges when you graph the function of the form y  x 2 is called
parabola.
a := x
2
b := ( x + 1 )
2
c := ( x - 1 )
2
2
d := 3 ( x + 1 ) + 2
2
e := 3 ( x - 1 ) - 2
2
p1 := x + 2 x + 1
2
p2 := x - 2 x + 1
2
p3 := 3 x + 6 x + 3
2
p4 := 3 x - 6 x + 3
The graph of the function y  a ( x  h ) 2  k is the same wide as the function y = ax^2 it
means the graph are congruent but their positions are different, because h and k moves
them to the right, left, up or down.
The coordinates of the parabola.
Ax + bx + c = a [ x^2 + b/a x ] + c
= a [( x + b/2a )^2 – b^2/4a^2 ] + c
= a ( x + b/2a )^2 – b^2/4a + c
= a ( x + b/2a ) + 4ac – b^2/4a
= a ( x – (- b/2a))^2 + 4ac – b^2/4a
Comparing with y = a(x – h)^2 + k gives h = - b/2a and k = 4ac – b^2/4a
PART 3-2
Though not so simple as the circle, the ellipse
is nevertheless the curve most often "seen" in
everyday life. The reason is that every circle,
viewed obliquely, appears elliptical.
Any cylinder sliced on an angle will
reveal an ellipse in cross-section (as
seen in the Tycho Brahe Planetarium
in Copenhagen).
Tilt a glass of water and the surface of the
liquid acquires an elliptical outline. Salami is
often cut obliquely to obtain elliptical slices
which are larger.
The early Greek astronomers thought that the planets moved in circular
orbits about an unmoving earth, since the circle is the simplest
mathematical curve. In the 17th century, Johannes Kepler eventually
discovered that each planet travels around the sun in an elliptical orbit
with the sun at one of its foci.
The orbits of the moon and of artificial
satellites of the earth are also
elliptical as are
the paths of comets in permanent
orbit around the sun.
On a far smaller scale, the electrons
of an atom move in an approximately
elliptical orbit with the nucleus at one
focus.
The ability of the ellipse to rebound an object starting from one focus to
the other focus can be demonstrated with an elliptical pool table. When a
ball is placed at one focus and is
thrust with a cue stick, it will rebound
to the other focus. If the pool table is
live enough, the ball will continue
passing through each focus and
rebound to the other.
Statuary Hall in the U.S. Capital building is elliptic. It was in this room that John Quincy
Adams, while a member of the House of Representatives, discovered this acoustical
phenomenon. He situated his desk at a focal
point of the elliptical ceiling, easily
eavesdropping on the private
conversations of
other House members located
near the other
focal point.
One of nature's best known approximations
to parabolas is the path taken by a body
projected upward and obliquely to the pull
of gravity, as in the parabolic trajectory of a
golf ball. The friction of air and the pull of
gravity will change slightly the projectile's
path from that of a true parabola, but in
many cases the error is insignificant.
This discovery by Galileo in the 17th
century made it possible for cannoneers
to work out the kind of path a
cannonball would travel if it were
hurtled through the air at a specific
When a baseball is hit into the air, it follows
angle.
a parabolic path; the center of gravity of a
leaping porpoise describes a parabola.
The opposite principle is used in the
giant mirrors in reflecting telescopes
and in antennas used to collect light and
radio waves from outer space: the beam
comes toward the parabolic surface
and is brought into focus at the focal point
The largest parabolic mirror in
existence is in a telescope located in the Caucasus
mountains in Russia. It is nearly 20 feet
in diameter and was completed in 1967.
The easiest way to visualize the path of
a projectile is to observe a waterspout.
Each molecule of water follows the
same path and, therefore, reveals a
picture of the curve.
Parabolas exhibit unusual and useful
reflective properties. If a light is placed at
the focus of a parabolic mirror (a curved
surface formed by rotating a parabola about
its axis), the light will be reflected in rays
parallel to said axis. In this way a straight
beam of light is formed. It is for this reason
that parabolic surfaces are used for
headlamp reflectors. The bulb is placed at
the focus for the high beam and in front of
the focus for the low beam.
Two types of images exist in nature: real
and virtual. In a real image, the light
rays actually come from the image. In a
virtual image, they appear to come from
the reflected image - but do not. For
example, the virtual image of an object
in a flat mirror is "inside" the mirror, but
light rays do not emanate from there.
Real images can form "outside" the
Heat waves, as well as light and sound
system, where emerging
waves, are reflected to the focal point of a
light rays cross
parabolic surface. If a parabolic reflector is
and are caught
turned toward the sun, flammable material
- as in a Mirage, an
placed at the focus may ignite. (The word
arrangement of two
"focus" comes from the Latin and means
concave parabolic
fireplace.) A solar furnace produces heat by
mirrors. Mirage's 3-D
focusing sunlight by means of a parabolic
mirror arrangement. Light is sent to it by set
illusions are
of moveable mirrors computerized to follow
similar to the holograms
the sun during the day. Solar cooking
formed by lasers.
involves a similar use of a parabolic mirror.
PART 3-3
ax
2
 bx  c  0
ax
2
 bx   c
ax
2
a
x
☼ There is a perfect relation
between the zeros of a
quadratic function y=f(x)
and the solution of the
quadratic equation f(x)=0.
But what is the solution of
the quadratic equation? It
is the value or values of x
that makes the function 0.
To find the solutions of the
quadratic equations we
need to find the formula for
it , by solving general
quadratic equation for x.
bx

a

2
bx
a
 
a
x
c
a
 b 
 2
x  

2a
 2a 
b
2
b 

x 

2
a


2
b 

x 

2a 

2
b 

x 

2a 

2

c
 
b
 

a
b

b

2

2
2
2
2
c
a
2
4a
b
4a
2
4a
 b 
 
 

a
 2a 
c

4 ac
4a
2
 4 ac
2
4a
x 
c
2
b
2
 
b
2
2a
x  
 4 ac
4a
b
2a

b
2
2
 4 ac
2a
 
b
2  4 ac
2a

 b 
b
2
2a
 4 ac
By using this quadratic formula we can find the solutions of the
ax
2
 bx  c  0
where
a0
. The discriminant which is the expression under the radical
b  4 ac
2
tells us how many solutions does the particular quadratic equation has
one, two or none.
If
If
b
2
 4 ac 
0 , the equation has two real solutions
b  4 ac  0 , the equation has one real solution
If b 2
2
 4 ac  0 , the equation has no real solution
2
2
p1 := x - 4 x - 5
p2 := -2 x + 5 x - 2
x  4x  5  0
2
 2x
a  1, b   4 , c   5
a   2 , b  5, c   2
b  4 ac    4   4 1   5   36
2
2
x
b
b  4 ac
2
2a
 x

  4  
2 1 
 5x  2  0
2
b
36

4
46
2
2
 4 ac  5 
36
2
x 
 5 
2  2 
 x 
 x5
 x  1
We found two solutions -1 and 5.
9
 x 
 4   2   2   9

 5  3
 4
 5  3
 4
2

 5  3
 4
1
2
 2
We found two solutions ½ and 2.
2
p3 := 3 x - 5
3x
2
2
p4 := 4 x + 4 x + 1
5  0
a  3, b  0 , c   5
b
2
4x
 4 ac   0   4 3   5   60
x 
 0
4  2
15
x 
3
.

 4x  1  0
a  4, b  4, c  1
60
2 3 
 x 
 x 
2
15
 4  4 1   0
 4 
2 4 
0
 
1
2
3
We found two solutions
15 / 3 and
 15 / 3
We found one solution.
2
p5 := 7 x + 4 x + 3
7x  4x  3  0
2
4  2
 4 7 3    68
 x 
4
 68
14
 68 is undefined in real numbers, so there
is no solution.
PART IV Conclusion
Once we start studying scenic section we get a new definition
of a parabola as the focus of point equidistant to a fixed point called
focus and a line called divectrix. From this definition it was constructed the equation
2
of the parabola we get among other 3 standard forms that: x = 4 cy was such equation
(check construction if you want. Here c is the focal distance (distance from the vertex to
2
the focus of the parabola) So, y = 1  x 2  is of the form y = ax (here a > 0)
4c
which coincide with the old quadratic function whose graph we use to call a parabola.
Now let’s prove it has the focal property we claim it has. Let’s take any point
( x0 , y0 ) in the graph of “ such” parabola
So if y =
1
4c
x
2
then Po = (x 0 , y0 ) = (x 0 ,
x0
2
4c
)
To define the angles of incidence and reflection we are using the angle formed with the
tangent line to the curve at F.
We need to prove that
P0 QF is isosceles and that the tangent line through P0 is
angle bisector of angle FP0 Q, in other words we need to prove that
P0 CQ = P0 CF
(congruent) which make

FP0 C = CP0 Q and sine

CP0 Q = incidence angle (vertical
angle) we are going to be done.
We need to find the equation of tangent line to check that this line intercept segment FQ
in the midpoint and than everything will follow to do so; we need the slope of the
tangent line so we need to borrow from Calculus I the geometrical interpretation
of the derivative of a function at a point
f(x) =
f(x) =
f(x 0 ) =
x
2
4c
2x
4c
x0
2c

x
2c
on slope tangent line
Belongs to tangent line
so, we have
0=
P0 (x0 , y0 ) and m =
y- y 0 =
y=
x0
2c
x0
2c
x
x0
2c
0=
(x - x 0 ) eq of tangent line.
 x0
200
2c
 y0 
x0
2c
x
x02
2
2c

x0
2
4c
x0
2c
x0
x0
x 2c -
2
4c
+
x0
x02
x0

2c
2
4c
2
-
x
2
4c
2
0
2c
=
Check Midpoint FQ = (
x
x0
2
2
0
-
2c
, 0)
x
2
0
2c
=0
THE END
THANK YOU