Solution
David answered on
Dec 22 2021
Introduction:
Mathematics is a huge section, there is no end. Each time mathematicians stuck at some
problem and think maybe this is the ultimate problem if we find out the solution to this problem
there won’t be any more hard problems. But each and every time one problem gets solved few
more come into picture. That is also a reason why mathematics is so challenging and interesting.
And due to this diversity and interesting nature it is the favorite topic of a large portion of
population. I, personally as a math lover, always try to learn different sections as well as this
subject more deeply. This is one of the important reason I completed this paper on a special topic
of Mathematics, “Shadow Functions”.
Now the “shadow function” is not only interesting but it has a great importance in
mathematics (especially in graph theory). For any function (whose order is higher than 1 i.e. a
quadratic or cubic or higher degree polynomial) there exists a function of somewhat same form
(that function is also a quadratic if the original function is quadratic and so on) which has the
same zeroes as the previous (original) one but with forms of opposite concavity. Moreover is the
original polynomial has zeroes in complex form then its shadow function will have zeroes in real
form. This special function has various examples in real life also, which I discussed at the right
place. For this Mathematics portfolio, I will be considering the shadow function of two given
form. I have two equation forms, the 1
st
one is a quadratic and the 2
nd
one is cubic. And this
paper will be discovering the facts related to these two functions, their shadow functions and in
general the shadow function of same degree equation. For that, in this paper, I will create two
portions; “Part A” will be for the considered quadratic form and “Part B” for cubic form. I will
e trying to find out the required relations in both the portions and I am sure the related graphs
are going to help me a lot to complete my work.
But most of the works and drawing of graphs are time consuming and hard. So to make
my work simple and easy I will take help of technology and computers. And due to need of
many graphs the best software I can think of is well-known software “GeoGe
a”. It is really
user friendly and has a nice and easy output which can be understood by any common man.
Part A (Quadratic Polynomials)
Here we have the equation,
y1 =(x-a)
2
+
2
; and we want to find out the coordinates of the vertex.
As we can see for any real constant value of b,
2
is positive. So the vertex of this
quadratic polynomial will depend on the other part.
Now,
y1 = (x-a)
2
is same as the equation y = x
2
(We can always consider z= (x-a) so thus the
statement) but shifted a units to the right. Since the vertex for y= x
2
is at (0, 0) and
consequently for equation y= (x-a)
2
vertex is at the point (a,0) , for our case the equation
also contains a constant term
2
thus here the vertex for this given equation is (a,
2
).
Now to show the zeroes of y1 consider the following part.
This second part can be proved logically, assuming b≠0 (i.e.
2
>0) we have seen that the
vertex of this equation is at (a, b²), which have a positive, non-zero, y value. Again the
coefficient of x
2
is positive implying that this is a positive parabola and thus its
anches
will point upwards from the vertex. Therefore, for any b≠0 this parabola never crosses or
even touches the x-axis, implying that it has no real zeros.
It can also be verified by calculation,
we expand the equation to mx²+nx+o form, then calculating D = n²-4mn is giving a
negative value meaning there are no real zeros (m=1,n= -2a and o=a
2
+
2
and for any non-
zero b D is -4
2
).
Now in order to calculate the zeros, we can use the formula, y =
√
=
√
which gives the zeroes as a ib.
Now the zeroes of the shadow function is a b which is giving the equation as,
y2 = -(x-a)
2
+
2
The question is how am I getting this equation?
Well, if we have zeros as α and β then,
(x-α)(x-β) = 0 which gives the form as,
x
2
–(α+β)x + αβ = 0
Here α = a+b and β=a-b and putting these values in above form simplifying the form will
lead to the above equation (we also have to remember its of opposite concavity thus there
will be a negative sign outside i.e. it will be multiplied with “-1”).
The required graphs can be easily obtained from “GeoGe
a”. We just need to enter the
equations of the quadratic form and its shadow function in “Geoge
a” with
co
esponding to a and b. And as we change the value of a and b the graph will change its
position as we can see in the following graphs....