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

Problem : Draw graph of $y=\sin \left|x\right|$ .

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 $x=\mathrm{0,}y=e^{0+1}=e$ . 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 $y=x^2-2\left|x\right|-3$

Solution : The given expression $f\left(x\right)=x^2-2\left|x\right|-3$ 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 $f\left(x\right)={\left|x\right|}^2-2\left|x\right|-3=x^2-2\left|x\right|-3$ , 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 :

$$\Rightarrow y=\frac{1}{\left|x\right|+1}$$

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

$$\Rightarrow y=\frac{1}{x+1}$$

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

$$\Rightarrow y=\frac{1}{x}$$

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 $h\left(x\right)=g\left(\left|x\right|\right)=1/\left(\left|x\right|+1\right)$ by removing left half of the graph and taking mirror image of right half of the graph in y-axis. .