# 4.2 Transformation of graphs by modulus function  (Page 3/4)

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From the point of construction of the graph of y=|f(x)|, we need to modify the graph of y=f(x) as :

(i) take the mirror image of lower half of the graph in x-axis

(ii) remove lower half of the graph

This completes the construction for y=|f(x)|.

Problem : Draw graph of $y=|\mathrm{cos}x|$ .

Solution : We first draw the graph of $y=\mathrm{cos}x$ . Then, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of $y=|\mathrm{cos}x|$

Problem : Draw graph of $y=|{x}^{2}-2x-3|$

Solution : We first draw graph $y={x}^{2}-2x-3$ . The roots of corresponding quadratic equation are -1 and 3. After plotting graph of quadratic function, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of $y=|{x}^{2}-2x-3|$

Problem : Draw graph of $y=|{\mathrm{log}}_{10}x|$ .

Solution : We first draw graph $y={\mathrm{log}}_{10}x$ . Then, we take the mirror image of lower half of the graph in x-axis and remove lower half of the graph to complete the construction of graph of $y=|{\mathrm{log}}_{10}x|$ .

## Modulus function applied to dependent variable

The form of transformation is depicted as :

$y=f\left(x\right)\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}|y|=f\left(x\right)$

As discussed in the beginning of module, value of function is first calculated for a given value of x. The value so evaluated is assigned to the modulus function |y|. We interpret assignment to |y| in accordance with the interpretation of equality of the modulus function to a value. In this case, we know that :

$|y|=f\left(x\right);f\left(x\right)>0\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}y=±f\left(x\right)\phantom{\rule{1em}{0ex}}$

$|y|=f\left(x\right);f\left(x\right)=0\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}y=0$

$|y|=f\left(x\right);f\left(x\right)<0\phantom{\rule{1em}{0ex}}⇒\text{Modulus can not be equated to negative value. No solution}$

Clearly, we need to neglect all negative values of f(x). For every positive value of f(x), there are two values of dependent expressions -f(x) and f(x). It means that we need to take image of upper part of the graph across x-axis. This is image in x-axis.

From the point of construction of the graph of |y|=f(x), we need to modify the graph of y=f(x) as :

1 : remove lower half of the graph

2 : take the mirror image of upper half of the graph in x-axis

This completes the construction for |y|=f(x).

Problem : Draw graph of $|y|=\left(x-1\right)\left(x-3\right)$ .

Solution : We first draw the graph of quadratic function given by $y=\left(x-1\right)\left(x-3\right)$ . Then, we remove lower half of the graph and take mirror image of upper half of the graph in x-axis to complete the construction of graph of $|y|=\left(x-1\right)\left(x-3\right)$ .

Problem : Draw graph of $|y|={\mathrm{tan}}^{-1}x$ .

Solution : We first draw the graph of function given by $y={\mathrm{tan}}^{-1}x$ . Then, we remove lower half of the graph and take mirror image of upper half of the graph in x-axis to complete the construction of graph of $y={\mathrm{tan}}^{-1}x$ .

## Modulus function applied to inverse function

The form of transformation is depicted as :

What is power set
Period of sin^6 3x+ cos^6 3x
Period of sin^6 3x+ cos^6 3x