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Negation of independent variable

The graph flipped across y. -axis.

The form of transformation is depicted as :

y = f x y = f - x

A graph of a function is drawn for values of x in its domain. Depending on the nature of function, we plot function values for both negative and positive values of x. When sign of the independent variable is changed, the function values for negative x become the values of function for positive x and vice-versa. It means that we need to flip the plot across y-axis. In the nutshell, the graph of y=f(-x) can be obtained by taking mirror image of the graph of y=f(x) in y-axis.

While using this transformation, we should know about even function. For even function. f(x)=f(-x). As such, this transformation will not have any implication for even functions as they are already symmetric about y-axis. It means that two parts of the graph of even function across y-axis are image of each other. For this reason, y=cos(-x) = cos(x), y = |-x|=|x| etc. The graphs of these even functions are not affected by change in sign of independent variable.

Problem : Draw graph of y=cosec(-x) function

Solution : The plot is obtained by plotting image of core graph y=cosec(x) in y axis.

Changing sign of the argument of graph

The transformed graph is image of core graph in y-axis.

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Combined input operations

Certain function are derived from core function as a result of multiple arithmetic operations on independent variable. Consider an example :

f x = - 2 x - 2

We can consider this as a function composition which is based on identity function f(x) = x as core function. From the composition, it is apparent that order of formation consists of operations as :

(i) f(2x) i.e. multiply independent variable by 2 i.e. shrink the graph horizontally by half.

(ii) f(-2x) i.e. negate independent variable x i.e. flip the graph across y-axis.

(iii) f(-2x-2) i.e. subtract 2 from -2x.

This sequence of operation is not correct for the reason that third operation is a subtraction operation to -2x not to independent variable x, whereas we have defined transformation for subtraction from independent variable. The order of operation for transformation resulting from modifications to input can, therefore, be determined using following considerations :

1 : Order of operations for transformation due to input is opposite to the order of composition.

2 : Precedence of addition/subtraction is higher than that of multiplication/division.

Keeping above two rules in mind, let us rework transformation steps :

(i) f(x-2) i.e. subtract 2 from independent variable x i.e. shift the graph right by 2 units.

(ii) f(2x-2) i.e. multiply independent variable x by 2 i.e. shrink the graph horizontally by half.

(iii) f(-2x-2) i.e. negate independent variable i.e. flip the graph across y-axis.

This is the correct sequence as all transformations involved are as defined. The resulting graph is shown in the figure below :

Graph of transformed function

Operations are carried in sequence.

It is important the way graph is shrunk horizontally towards origin. Important thing is to ensure that y-intercept is not changed. It can be seen that function before being shrunk is :

f x = x - 2

Its y-intercept is 2. When the graph is shrunk by a factor by 2, the function is :

f x = 2 x - 2

The y-intercept is again 2.The graph moves 1 unit half of x-intercept towards origin. Further, we can verify validity of critical points like x and y intercepts to ensure that transformation steps are indeed correct. Here,

x = 0, y = - 2 X 0 2 = - 2 y = 0, x = - y + 2 2 = - 2 2 = - 1

We can decompose a given function in more than one ways so long transformations are valid as defined. Can we rewrite function as y = f{-2(x+1)}? Let us see :

(i) f(2x) i.e. multiply independent variable x by 2 i.e. i.e. shrink the graph horizontally by half.

(ii) f(-2x) i.e. negate independent variable i.e. flip the graph across y-axis.

(iii) f{-2(x+1} i.e. add 1 to independent variable x x i.e. shift the graph left by 1 unit.

This decomposition is valid as transformation steps are consistent with the transformations allowed for arithmetic operations on independent variable.

Graph of transformed function

Operations are carried in sequence.

Horizontal shift

We have discussed transformation resulting in horizontal shift. In the simple case of operation with independent variable alone, the horizontal shift is “c”. In this case, transformation is represented by f(x+c). What is horizontal shift for more general case of transformation represented by f(bx+c)? Let us rearrange argument of the function,

f b x + c = f { b x + c b }

Comparing with f(x+c), horizontal shift is given by :

Horizontal shift = c b

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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