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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 :

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 = 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.

Modulus operator applied to sine function

Modulus operator applied to the argument of sine function.

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.

Modulus operator applied to exponential function

Modulus operator applied to the argument of exponential function.

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 x = 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 x = x 2 - 2 x - 3 . Hence, we first draw f x = x 2 - 2 x - 3 . The corresponding quadratic equation f x = x 2 - 2 x - 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 x = | 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.

Modulus operator applied to quadratic function

Modulus operator applied to the quadratic function.

Problem : Draw graph of function defined as :

y = 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 = 1 x + 1

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

y = 1 x

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

Modulus operator applied to rational function

Modulus operator applied to the argument of rational function.

Modulus function applied to the function

The form of transformation is depicted as :

y = f x y = | f x |

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.

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply

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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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