Archimedes of SyracuseArchimedes of Syracuse (Circa 287 BC - 212 BC), was a Greek mathematician, astronomer, philosopher, physicist and engineer. He was killed by a Roman soldier during the sack of the city, despite orders from the Roman general, Marcellus, that he was not to be harmed. (The Greeks said that he was killed while drawing an equation in the sand, and told this story to contrast their high-mindedness with Roman ham-handedness; however, it should be noted that Archimedes designed the siege engines that devastated a substantial Roman invasion force, so his death may have been out of retribution.) Some math historians consider Archimedes to be one of history's greatest mathematicians, along with possibly Newton, Gauss, and Euler.

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Light is known to behave in a very predictable manner. If a ray of light could be observed approaching and reflecting off of a flat mirror, then the behavior of the light as it reflects would follow a predictable law known as the law of reflection.

1.b1.b

1.a1.a

ArchimedesArchimedes constructed a kind of hexagonal mirror, and at an interval proportionate to the size of the mirror he set similar small mirrors with four edges, moved by links and by a form of hinge, and made it the centre of the sun's beams--its noon-tide beam, whether in summer or in mid-winter. Afterwards, when the beams were reflected in the mirror, a fearful kindling of fire was raised in the ships, and at the distance of a bow-shot he turned them into ashes.

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1.c1.c

2.a2.a In my opinion plane mirror is known to behave in a very predictable manner. If a ray of light could be observed approaching and reflecting off of a flat mirror, then the behavior of the light as it reflects would follow a predictable law known as the law of reflection. The diagram illustrates the law of reflection. And defiantly would not affect as Hexagonal mirror. 2.b2.b No, It would have the same properties, because it has different shape and has different way of use.

For example, today we use plane mirrors as a reflection of ourselves and Hexagonal mirrors used in surgery rooms and as a microscopic devices.

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2.c2.c I don’t think so, because we do not get burned from plain mirrors.

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Question 1a.  Find the Vertex of fallowing function :

Answer 1a.

We can see that the formula we should use to solve this quadratic function is .

 Now we solve                          

After we got our Vertex we determine if the it is a Maximum or Minimum Point

Since a = -1, and -1 is less than 0, this parabola would open down  .

So our vertex (- 4, -2) is the maximum point.

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Question 1b.  Find the Vertex of fallowing function :

Answer 1b.

We can notice that the fallowing function written as

So the formula we should use to solve this quadratic function is .

After we got our Vertex we determine if the it is a Maximum or Minimum Point

Since a = 1, and 1 is greater than 0, this parabola would open down  .

So our vertex (1, 1) is the minimum point.

Now we Identify the values as:

And plot them

Now we plug the x to find the y

The vertex would be (1, 1).

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Question 2a.  Find the Vertex of fallowing function :

Answer 2a. We can see that the formula we should use to solve this quadratic function is

So our vertex is (-2, 1) and in the minimum point.Now we will find the x, y intercepts by replacing them with 0

X=0 Y=0

y-intercept is (0, 5).

Since the square root is negative we can not find the x-intercept

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After we got our Vertex we determine if it is

a Maximum or Minimum Point. Since a = -2, and 1 < 0,

this parabola would open up  .

We solve

Question 2b.  Find the Vertex of fallowing function :

Answer 2b. We can see that the formula we should use to solve this quadratic function written as:

So our vertex (0, 1) is the minimum point.

We use the formula

        

Now we will find the x, y intercepts by replacing them with 0X=0 Y=0

y-intercept is (0, 1).

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After we got our Vertex we determine if the it is a Maximum or Minimum Point. Since a = 0, and 1 is less than 0, this parabola would open down  .

                   .                     

Since the x-intercepts are (-1, 0) and (1, 0).

We solve

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1.a1.a The name of plane curve that emerges the functions y=ax^2, a = 0 is Parabolic curve.

1.b1.b In my opinion it related next way:

1. Both share the same shape, which changes only if the values of a were different in both equations.

Click here to see better example in

Maple Illustration.

Invariant under certain DilationInvariant under certain DilationParabola have the property that when scaled (stretching/shrinking) along a direction parallel or perpendicular to its axis, the curve remain unchanged. (For example, line also have this property, but circle do not. A stretched line is still a line, but a stretched circle is no longer a circle) When a parabola is stretched along the direct x a units and along the axis by b units, the resulting curve is the original parabola scaled in both direction by a^2/b.2.b2.b Given a parameterization of a parabola {x f(t), y f(t)} with vertex at Origin and focus along the y-axis, it's focus is {0, xf(t)^2/[4y f(t)] }.Optical PropertyOptical PropertyA radiant point at the focus will reflect or refract off the parabola into parallel lines. The figure shows three parabolas, two of which share a common focus.

2.a2.a We use parabolas in everything that relates to curve and second power such as a cup, a bowl , and car lights as shown downward.

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Discriminate

3.a3.a

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First of all, when we see functions as: y=f(x) or f(x)=0 we know that the relation is: Discriminate=0

We use the standard form Where a= 0When a quadratic equation is in standard form, the expression that is found under the square root part of the quadratic formula is called the discriminate.  The discriminate can tell you how many solutions there are going to be and if the solutions are real numbers or complex imaginary numbers.

Two distinct real solutionsNote that the value of the discriminate is found under the square root and there is a + or - in frontof it.  So, if that value is positive, then there would be two distinct  real number answers

Kinds of solution

for

1) If

One real solution

Note that the value of the discriminate is found

under the square root and there is a + or - in front

of it. So, if that value is zero, + or - zero is the

same number, so there would be only one

real number solution.

2) If

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3.a3.a

3) If

Two distinct complex imaginary solution

Note that the value of the discriminate is found

under the square root and there is a + or - in front of it.

 So, if that value is negative, then there

would be two distinct complex imaginary

number answers.

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44 Real coefficients in a second-degree polynomial arecrucial keys to determine and draw the graph of the function.To find out vertex, and whether the graph curve up or down, real coefficients give us the image of the graph of function.

f (x) =ax^2 (a) is a positive real coefficient in a second-degree polynomial, this gives us information, which is the direction of the curve. In the general case, f (x) =ax^2 + bx + c , we find the coordinates of the vertex of the parabola by using the formula: [-b/2a, f (-b/2a)], a, and b, are keys to find the vertex, and c, is the y-intercept.

For example:

In this case, the property of (a) is same as a in the first function. As a result, it is a reflection over the x- axis. The real coefficient a is an integral key to form the shape of the graph of the function. Moreover, it depends on it, and the parabola varies wider or narrower.

Another function is: f (x) = -ax^2

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Yes, I think that all of the quadratic equations has focal property, because conic functions sections can be defined as curves such that the ratio of the distances from a point on the curve to a straight line (the directrix) and a point (the focus) is a constant, called the eccentricity e.

This is called the focal definition. If the directrix is taken as the y-axis, and the focus is at (p,0), then this definition gives [(x - p)2 + y2]1/2/x = e. Multiplying out,

x2(1 - e2) + y2 - 2px + p2 = 0. It's easy to see that this quadric is an ellipse if e < 1. a hyperbola if e > 1, and a parabola if e = 1. The ellipse and hyperbola are called central conics, for which the origin is usually taken at the centre. If this is done by making the substitution x = x' + h in the equation, then h = p/(1 - e2). For an ellipse, this puts the directrixes at x' = ±(a/e); that is, on the two sides. As these directrixes recede to infinity, the ellipse becomes a circle. We can show that e = c/a, which is the usual definition of the eccentricity of an ellipse. Here, a is the semi major axis and c is the focal distance. The directrixes of a hyperbola are between the branches, again at distances ±(a/e), but here e > 1. The focal definitions of the ellipse and hyperbola are not of much use in applications, in contrast with the focal definition of the parabola.

Paraboloids are found very commonly as reflectors of waves, in satellite antennas ("dishes"), searchlights and listening devices. They bring any kind of waves to a focus, whether light, radio waves, sound, or even water waves. A bay with a parabolic beach will concentrate waves at the focus. I do not know of any examples of this, but similar effects of ocean wave refraction are well known. All of these uses depend on the focal property. A spherical concave mirror also has this property, but it is only approximate. To do better than a spherical mirror, a paraboloid must be rather accurate. In small sizes, it is easier to make an accurate spherical mirror than an accurate paraboloid, and it is also better than a bad paraboloid. The paraboloid must be accurately pointed to profit from its advantages. Otherwise, a spherical mirror is actually superior.

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Professor : Orlando Alonso

Student: Elyahu Pinhasov Course : Math 200

> a:=3*x^2;h:=3*(x+2)^2;j:=3*(x+2)^2-5;

a := 3 x 2

j := 3 x + 2( )2 - 5

> plot([a,h,j],x=-5..3,y=-5..27);

As we can see, the functions in front of us have congruent geometrical representations. They represent vertical and horizontal shift of each other.

h: = 3(x +4x+4) h: = 3x +12x+12h := 3 x + 2( )2 2 2

h: = (-2,0)

j: = 3(x +4x+4) -52

j: = 3x +12x+12-52

j: = 3x +12x+72

> a:=3*x^2;

a := 3 x 2

> plot([a],x=-3..3,y=-5..27);

> b:=-3*x^2;

b := -3 x 2

> a:=3*x^2;plot([a,b],x=-3..3,y=-27..27);

a := 3 x 2

> a:=3*x^2;c:=3*x^2+12*x+12;d:=3*x^2+12*x+7;

a := 3 x 2

c := 3 x 2 + 12 x + 12

d := 3 x 2 + 12 x + 7

> plot([a,c,d],x=-5..3,y=-5..27);