# 4.2 Transformation of graphs by modulus function  (Page 2/4)

 Page 2 / 4

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 left half of the graph

2 : take the mirror image of right half of the graph in y-axis

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

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

Solution : First we draw graph of sinx. In order to obtain the graph of y=sin|x|, we remove left half of the graph and take the mirror image of right half of the graph of in y-axis.

Problem : Draw graph of $y={e}^{|x+1|}$ .

Solution : We first draw graph of $y={e}^{x}$ . Then, we shift the graph left by 1 unit to obtain the graph of ${e}^{x+1}$ . At $x=0,y={e}^{0+1}=e$ . In order to obtain the graph of $y={e}^{|x+1|}$ , we remove left part of the graph and take the mirror image of right half of the graph of $y={e}^{x+1}$ in y-axis.

In order to obtain the graph of $y={e}^{|x+1|}$ , we remove left part of the graph and take the mirror image of right half of the graph of $y={e}^{x+1}$ in y-axis.

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

Solution : The given expression $f\left(x\right)={x}^{2}-2|x|-3$ is obtained by taking modulus of the independent variable of the corresponding quadratic polynomial in x as given here, $f\left(x\right)={x}^{2}-2x-3$ . Hence, we first draw $f\left(x\right)={x}^{2}-2x-3$ . The corresponding quadratic equation $f\left(x\right)={x}^{2}-2x-3=0$ has real roots -1 and 3. The co-efficient of “ ${x}^{2}$ ” is positive. Hence, its plot is a parabola which opens upward and intersects x-axis at x=-1 and x=3.

In order to draw the graph of $f\left(x\right)={|x|}^{2}-2|x|-3={x}^{2}-2|x|-3$ , we remove left half of the graph and take the mirror image of right half of the core graph of quadratic function in y-axis.

Problem : Draw graph of function defined as :

$⇒y=\frac{1}{|x|+1}$

Solution : It is clear that we can obtain given function by applying modulus operator to the independent variable of function given here :

$⇒y=\frac{1}{x+1}$

This function, in tern, can be obtained by applying shifting modification to the argument of the function given as :

$⇒y=\frac{1}{x}$

We, therefore, first draw $f\left(x\right)=1/x$ . Then we draw $g\left(x\right)=f\left(x+1\right)=1/\left(x+1\right)$ by shifting the graph left by 1 unit. Finally, we draw $h\left(x\right)=g\left(|x|\right)=1/\left(|x|+1\right)$ by removing left half of the graph and taking mirror image of right half of the graph in y-axis. .

## Modulus function applied to the function

The form of transformation is depicted as :

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

It can be seen that modulus operator here modifies the value of the function itself. In other words, it is like changing output of the function in accordance with nature of modulus function. The output of the function is now either zero or positive number. This has the implication that part of the graph y=f(x) corresponding to negative function values is not present in the graph of y=|f(x)|. Rather, negative function value of f(x) is converted to positive function value. This change in the sign of function takes place without changing magnitude of the value. It implies that we can obtain function values, which correspond to negative function value in y=f(x) by taking image of negative function values across x-axis. This is image in x-axis.

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