Math Hl Type 1 Shadows

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  • Topic: Quadratic equation, Polynomial, Complex number
  • Pages : 12 (3091 words )
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  • Published : January 27, 2013
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Higher Level Mathematics
Internal Assessment Type I
Shadow Functions


Introduction: Functions/Polynomials3
Part A: Quadratic Polynomials4
Part B: Cubic Polynomials12


In mathematics, function is defined as a relationship, or more of a correspondence between the set of input values and the set of output values. Also, a rule is involved, or as it may be referred to, a ‘set of ordered pairs’ that assigns a unique output for each of the input. The output correspondence is usually defined as f and the output is x. The correspondence is denoted as f(x). All functions are mainly defined by two factors, as was mentioned before, set of inputs - which are called arguments; and outputs - which are oftenly called values. The set of all arguments is called domain; and the set of all the values is called range.

The graph on the left is just an example of a simple function f(x)=3x+2. As you can see it is a straight line.

Any function has an ability to be described through its relations to other functions, for example as an inverse function, or as a solution to a differential equation. Also, as we will see further, functions can be quadratic. Quadratic function graph is simply a parabola. The equation of that function doesn’t contain any powers that are higher than 2. Basically, quadratic function is a polynomial of degree of 2.

There are different types of polynomials, but the most common ones and with what we are going to be working further, quadratic and cubic type of functions with polynomial equations of e.g. x2-2x+12 or x3-2x2-3x+13 etc.

Also, there is another type of functions, The Shadow Functions, which aren’t commonly used in everyday life, though still do deserve attention. Generally, Shadow Function is a simple reflection of an original function but via shadow generating function that shares the same vertex as the original and the shadow functions.Those are going to be investigated further in this report.

Part A (Quadratic Polynomials)

Consider the quadratic function Y1 = (x-a)2+b2 where a and b ∈ℜ

Here, we are given a quadratic function - parabola, which is written in a vertex form. Vertex is the highest or the lowest point of the parabola (depending on whether the parabola is opened up or down).

Using the equation that we are provided with, the vertex of the function here would be (a; b2), where a is taken with an opposite sign and b, taken as it is.

In order to prove that this particular function has zeros of a±ib, it is required to make Y1 equal to zero. Then find the x ‘s of the function using the quadratic formula: x=-b±b2-4ac2a

Y1 = (x-a)2+b2
(x-a)2+b2= 0
x2-2xa+a2+b2= 0
a = 1
b = -2a
c = (a2+b2)

x=2a±(4b2×-1)2i = -1
x=2a2±i4b22= a±i4b22= a±2b2i= a±bi

As you can see, it has been proven that function Y1 = (x-a)2+b2 does really have zeros: a±bi
The “shadow function” to Y1 is another quadratic function Y2 which shares the same vertex as Y1. However, Y2 has opposite concavity to that of Y1 and its zeros are in the form a±b. Also, we are provided with “shadow generating function” Ym which mainly just a straight horizontal line, which passes through the vertex of Y1 and Y2.

In order to generate pairs of functions Y1 and Y2, I will substitute various values for a and b. The most sophisticated way would be to take different values in terms of their sign; type - rational, irrational (Π, e); and fractions. This, is to check whether the quadratic equations given are to be used with all of the values.

Y1 = f(x) = (x-a)2+b2
Y2 = g(x) = -(x-a)2+b2

a = 0Y1 = x2
b = 0Y2 = -x2

a = 2Y1 = (x-2)2+9
b = -3Y2 = -(x-2)2+9

a = -2Y1 = (x+2)2+9
b = 3Y2 = -(x+2)2+9

a = -13Y1 = (x+13)2+449
b = -27Y2 = -(x+13)2+449

a = ΠY1 = (x-Π)2+e2...
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