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Question Number 208031 by Davidtim last updated on 02/Jun/24
we have y=f(x) function, if we transfer  its graph C unit vertically and gain  the new function y=f(x)±C, it′s meant  y is increased or decreased  C units.  if we transfer the graph of considered  function horizontally, we gain the new function  y=f(x±C), what does mean it?
$${we}\:{have}\:{y}={f}\left({x}\right)\:{function},\:{if}\:{we}\:{transfer} \\ $$$${its}\:{graph}\:{C}\:{unit}\:{vertically}\:{and}\:{gain} \\ $$$${the}\:{new}\:{function}\:{y}={f}\left({x}\right)\pm{C},\:{it}'{s}\:{meant} \\ $$$${y}\:{is}\:{increased}\:{or}\:{decreased}\:\:{C}\:{units}. \\ $$$${if}\:{we}\:{transfer}\:{the}\:{graph}\:{of}\:{considered} \\ $$$${function}\:{horizontally},\:{we}\:{gain}\:{the}\:{new}\:{function} \\ $$$${y}={f}\left({x}\pm{C}\right),\:{what}\:{does}\:{mean}\:{it}? \\ $$
Answered by Frix last updated on 03/Jun/24
y=f(x); C>0  y+C=f(x) shifts down  y−C=f(x) shifts up  y=f(x+C) shifts left  y=f(x−C) shifts right
$${y}={f}\left({x}\right);\:{C}>\mathrm{0} \\ $$$${y}+{C}={f}\left({x}\right)\:\mathrm{shifts}\:\mathrm{down} \\ $$$${y}−{C}={f}\left({x}\right)\:\mathrm{shifts}\:\mathrm{up} \\ $$$${y}={f}\left({x}+{C}\right)\:\mathrm{shifts}\:\mathrm{left} \\ $$$${y}={f}\left({x}−{C}\right)\:\mathrm{shifts}\:\mathrm{right} \\ $$
Commented by Davidtim last updated on 03/Jun/24
I know it but you read my question again.
$${I}\:{know}\:{it}\:{but}\:{you}\:{read}\:{my}\:{question}\:{again}. \\ $$
Commented by Frix last updated on 03/Jun/24
Ok, I read it again. What now? Where′s  the point?
$$\mathrm{Ok},\:\mathrm{I}\:\mathrm{read}\:\mathrm{it}\:\mathrm{again}.\:\mathrm{What}\:\mathrm{now}?\:\mathrm{Where}'\mathrm{s} \\ $$$$\mathrm{the}\:\mathrm{point}? \\ $$

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