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   transformation of graphs 2
posted on 2 Jul 2007 11:47:57 IST    937 views    5 comments
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Tagged with:   20 Nickels awarded!
Case 4
To draw the graph of form y = | f(x) | (graph 4)
 
Draw the graph of y = f(x) (in yellow) then restore that part of the curve, which is above x-axis. Take the reflection of the part of the curve which is below x-axis (in yellow) wrt to x-axis. Then erase the part which is below x-axis... The restored part along with the reflection will be the required curve (in red dotted)
 
Case 5
To draw the graph of form y = f(|x|) (graph 5)
 
Draw the graph of y = f(x) (in yellow) and then erase the part which is to the left of y-axis. Take the reflection to the part of the curve lying to the right of y-axis assuming y-axis as mirror. The part of the curve, which is at the right of y-axis along with its reflection, is the required curve (in red dotted)
 
Case 6
To draw the graph of form y = f(- x) (graph 6)
 
Draw y = f(x) (in yellow)..and then take the total reflection of this graph wrt y-axis (in red)
 
Case 7
To draw the graph of form y = - f(x) (graph 7)
 
Draw y = f(x) (in yellow)..and then take the total reflection of this graph wrt x-axis (in red)




About the Author:
Manasi (3976)

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Olaaa!! Perrrfect answer.  702  bad job dude!! I dont approve of this answer!  2  [939 rates]

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  this article:   20 points  (with Olaaa!! Perrrfect answer.   in 4   votes   )     [?]
 
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SINDHURA KADIYALA is offline comment by SINDHURA KADIYALA      (posted on 2 Jul 2007 14:09:47 IST)
    its great........
Manasi is offline comment by Manasi      (posted on 3 Jul 2007 10:20:11 IST)
    i guess no one has to comment anything on this... no probs... jus make the best possible use of it!!!
joy francis is offline comment by joy francis      (posted on 3 Jul 2007 13:15:43 IST)
    thanks, this would quiet a help at the time of revision.
ac is offline comment by ac      (posted on 5 Jul 2007 23:50:57 IST)
    Hi, you've done pretty impressive work!
AC (Forum expert, IIT Delhi alumnus).
Manasi is offline comment by Manasi      (posted on 6 Jul 2007 12:48:09 IST)
    thank u sir!
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