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u represents any data point on the x-axis in the domain (0-7.5) which stand for the upper curve
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f(u) is the image for all the values of x and is permitted in a range of 2.5- 6.6
a graph has been drown to display the data points shown in table 1 and 2
The data points in the tables have been applied to model functions that follow the behavior of each curve shown. We can notice from the graph that these two curves are parallel to each other hence they might follow the same type of equation but vary in the parameters values.
Technology (GDC calculator) has been used to plot these curves and many types of functions have been examined to this model. I have come to the conclusion that the functions type that models the data behavior is a quartic equation.
Quartic equation = ax4+bx3+cx2+dx+ e
A quartic function is a symmetrical polynomial function because it is constructed from number of variables and constants, using only the operations of addition, subtraction, multiplication constant positive whole number exponents.
The reason I chose this function is because it is the most plausible function that fits and follows the behavior of the data points.
The two of the curves follow this function type but they differ in the parameters.
The parameters for the first curve (lower curve) are:
- a=0.018450285
- b=-0.898140257
- c=0.7355801435
- d=-1.013678057
- e=0.7300506358
Hence the function of the lower curve is:
f(x)= 0.018450285 x4 -0.898140257 x3 +0.7355801435 x2 -1.013678057x +0.7300506358
The parameters for the second curve (upper curve) are:
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a=-5.602082x10-4
- b=-0.0918509572
- c=1.096110546
- d=-2.471844394
- e=2.511349345
Therefore the function of the upper curve is:
f(u)= -5.602082x10-4u4 -0.0918509572u3 + 1.096110546u2-2.471844394u+2.511349345
There are some limitations that might not let these functions fit in perfectly. One of these limitation is we don’t have a wide range of values which means we are not able to state clearly if this function is symmetrical or not. Second because of the limited set of values and hence the limited domain and range, one of the curves may not be parallel to the other one and this makes on of the function wrong.
Logan wishes to print T-shirt with the logo on the back. She must double the dimensions of the logo. In order to achieve Logan’s whishes the loge must be stretched to the double from each of the dimension; the width and the length or height. Since the functions of the curves that surrounds the logo are know, it is now possible to apply function graphing skills on the these graphs or curves to be stretched.
In order to stretch a graph to the double parallel to the y-axis the formula must be multiplied by 2 for example:
The original equation: stretched equation:
f(x)= x2 2 f(x)=2x2
In order to stretch a graph to the double parallel to the x-axis the formula must be written this way
The original equation: stretched equation:
f(x)= x2 f(x)= ()2
Applying these techniques to our function:
f(x)= ax4+bx3+cx2+dx+ e
f(x)= 2[a()4+b()3+c()2+d+ e]
Each equation of the curves must be modified in this way in order to achieve the purpose of doubling the logo.
Logan’s also wishes to print business card. A standard business card has the dimension of 9 cm and 5 cm. The Original Square has the dimensions of 7.5cm by 7 cm. To stretch and squeeze it so it can fit in with the business cards dimensions, we need to know the ratio between the business card dimensions and the original square once.
5/7 = 0.7142857143
9/7.5= 1.2
So now if we multiply 7cm by 0.7142857143 the height of the original square will be compressed to 5cm. and if 7.5m is multiplied by 1.2 the length of the original square will be stretched up to 9cm.
With the purpose of extending the logo from one end of the card to the other, each function must be modified in a certain way. At first, we need to know to what extent the logo must be compressed and stretched. Then the function graphing skills can be employed. The amplitude of the logo in the original square is 6.2 cm this means that the logo is occupying 88% height of the square. Hence it should be compressed to 4.4286 cm. if we take the ratio of the two dimensions;
6.2/4.4286 = 0.7142857143 which the same as 5/7
The ratio of the length dimensions to stretch the logo is: 9.7/5 = 1.2
This means:
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the graph of f(x) and f(u) must be stretched parallel to the x-axis by a scale factor of 9.7/5= 1.2
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the graph of f(x) and f(u) must be stretched parallel to the y-axis by a scale factor of 5/7= 0.7142857143
If we indicate 5/7 with a symbol w, and to 0.833333 (1/1.2) with a symbol z. and apply this to our application f(x)= ax4+bx3+cx2+dx+ e:
w.f(zx)=w[a(zx)4+b(zx)3+c(zx)2+dzx+ e] = wa(zx)4+wb(zx)3+wc(zx)2+wdzx+ we
Business card with the logo extended on it
The fraction area can be measured by measuring the area under each of the curves and subtracting the area under the upper curve fron the area under the lower curve. This is how the area of the logo can be calculated
To measure the area under each of the curves the definite integral is taken for each of the graphs of the functions f(x) and f(u) where there is a lower limit and upper limit.
The equation of the lower curve in the card is:
f(x)= 0.0013178775(0.83333x)4 – 0.0637028755(0.83333x)3 +0.5254143882(0.83333x)2 – 0.724055755(0.83333x) + 0.5214647399=0
The equation of the upper curve in the card is:
f(u) = – 4.00148714x10-4(0.83333u)4 – 0.656078266(0.83333u)3 + 0.7829361043(0.83333u)2 – 1.765631742(0.83333u) + 1.793820961 = 0
the general equation after integration:
*Where the constant b of the integral in the lower limit
*the constant a of the integral is the upper limit
The area under the lower graph is:
= 10.74054666cm2
The area under the upper graph is:
= - 174.21259 < there is probably something wrong with the calculating
The accuracy of area of the logo is important in a business card because there are always some code and numbers written within these logos.
