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$$y=4\mathrm{sin}x$$
The amplitude of function "4sinx" is 4 times that of core graph "sinx". In the same fashion, a division by a positive constant greater than 1 results in shrinking of core graph by the factor, which is equal to constant being multiplies. Let us consider division of function :
$$y=\frac{1}{2}\mathrm{sin}x$$
The amplitude of the graph "sinx" changes from 1 to 1/2 in the graph of "1/2 sinx".
What would happen if we negate output of a function? Answer is easy. All positive values will turn negative and all negative values will turn positive. It means that graph of core function which is being negated will be swapped across x-axis in the transformation. The graph of “f(-x)”, therefore, is mirror image in x-axis. In other words, we would need to flip the graph f(x) across x-axis to draw graph “-f(x)”.
Problem : Draw graph of $y={\mathrm{log}}_{e}\frac{1}{x}$ .
Solution : We simplify given function as :
$$\Rightarrow y=\mathrm{log}{}_{e}\frac{1}{x}=\mathrm{log}{}_{e}1-\mathrm{log}{}_{e}x=-\mathrm{log}{}_{e}x$$
Here, core function is $\mathrm{f(x)}=\mathrm{log}{}_{e}\mathrm{(x)}$ . Clearly, given function is transformed function of type y=-f(x). We obtain its graph by taking mirror image of the graph of y=f(x) about x-axis. We obtain its graph by taking mirror image of graph of core function about x-axis.
Certain functions are derived from core function as a result of multiple arithmetic operations on the output of core function. Consider an example :
$$f\left(x\right)=-2\mathrm{sin}x-1$$
We can consider this as a function composition which is based on sine function f(x) = sinx as core function. Here, sequence of operations on the function is important. Difference in interpreting input and output composition is that input composition is evaluated such that defining input transitions are valid. This results in a order of evaluation which gives precedence to addition/subtraction over multiplication/division. This evaluation order is clearly opposite to normal composition order of arithmetic operations in which multiplication/division is given precedence over addition/subtraction. We, therefore, say that decomposition of function for input operation is opposite to that of composition order. In the case of output operation, however, composition order of arithmetic operations is maintained during decomposition. It is logical also. After all, we are operating on a value – not something that goes into function to generate values in accordance with function rule as is the case with independent variable. It is, therefore, expected that we carry out arithmetic operations on the function just the way we evaluate algebraic expressions. In the nutshell, we shall give precedence to multiplication/division over addition/subtraction. In the example abvoe, we subtract "-1" to "-2sinx" - not to core function "sinx".
Keeping above in mind, the correct sequence of operation for graphing is :
(i) 2f(x) i.e. multiply function f(x) by 2 i.e. stretch the graph vertically by 2.
(ii) -2f(x) i.e. negate function f(x) i.e. flip the graph across x-axis.
(iii) -2f(x) – 1 i.e. subtract 1 from -2f(x) i.e. shift the graph down by 1 units.
The combined input and output operation is symbolically represented as :
$$af\left(bx+c\right)+d;\phantom{\rule{1em}{0ex}}a,b,c,d\in R$$ Carrying out output operation before input operation does not make sense. There will be two different outputs which are not connected to each other. Hence, logical order is that we first carry out input operations then follow it with output operations.
Problem : Draw $y-1={\mathrm{log}}_{e}\left(x-2\right)$
Solution : We rewrite the function :
$$\Rightarrow y={\mathrm{log}}_{e}\left(x-2\right)+1$$
In order to plot this function, we plot the graph of core function $y={\mathrm{log}}_{e}x$ . Note that when y=0,
$$y={\mathrm{log}}_{{e}^{x}}=0\phantom{\rule{1em}{0ex}}\Rightarrow x={e}^{0}=1$$
In this case, plot intersects x-axis at x=1. Now, the plot of $y={\mathrm{log}}_{e}\left(x-2\right)$ is plot of $y={\mathrm{log}}_{e}x$ shifted right by 2 units. Note that when y=0,
$$y=\mathrm{log}e\left(x-2\right)=0\phantom{\rule{1em}{0ex}}\Rightarrow x-2={e}^{0}=1\phantom{\rule{1em}{0ex}}\Rightarrow x=3$$
The plot of $y={\mathrm{log}}_{e}\left(x-2\right)+1$ is plot of $y={\mathrm{log}}_{e}\left(x-1\right)$ shifted up by 1 unit.
There is yet another alternative to obtain graph of transformed function by shifting axes themselves instead of plot. In the case of shifting either in x or y direction, the operation of shifting graph is equivalent to shifting of axis. Therefore, transformation involving shifting can be affected by shifting axes in opposite directions to that required for the graph. In the example case, we need to move y-axis by 2 units towards left and move x-axis by 1 unit downwards.
Problem : Draw the plot $y={\mathrm{cos}}^{2}x$ .
Solution : We know that :
$$\Rightarrow y={\mathrm{cos}}^{2}x=\frac{1+\mathrm{cos}2x}{2}=\frac{1}{2}+\frac{\mathrm{cos}2x}{2}$$
Here, core graph is $y=\mathrm{cos}x$ . Multiplying independent variable by 2 shrinks core graph horizontally. As a result its period is reduced from 2π to π as shown in the graph. Division of cos2x by 2 is division operation on function. This operation shrinks the graph cos2x by 2 vertically. Note that amplitude of graph is reduced to 1/2 due to this operation. In the figure, lower graph corresponds to (cos2x)/2. Once we draw graph of (cos2x)/2, we draw given function $y={\mathrm{cos}}^{2}x$ by shifting the graph of (cos2x)/2 by 1/2 units up.
Author wishes to thank Ms. Aditi Singh, New Delhi for her editorial suggestions.
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