Pre-calculus question
It's either two units right, r two units left, Im on that section o math aswell XD when it's inside the parent function isn't it the opposite?
You have four sections left is negative and down is negative while up and right are positive
\[\rm f(x-h)\] here h is horizontal shift f(x) + k here k is vertical shift \[\huge\rm y =\left| x-h \right|+k\] if h is negative graph will shift h unit to the right if h is positive graph will shift h unit to the left if k is positive graph shift k unit p if k is negative some unit down
So what would be x-2
why two units right ? why not left ???
yes right. +h to the left -h to the right
If y = x-2 then the x intercept would be at x = 2 y = 0 , hence moving the graph to the right. If y = x + 2 , the x intercept would be at x = -2 , y = 0 , therefore moving to the left. You can graph it out to see the difference. c:
why do you think its A ??
if h is positive then graph shift to the `left` side if h is negative then graph shifts to the `right` side
@vera_ewing But h is minus in this case?
right. one unit to the right side
it's 2** not one
which one do you think ? you just said it or was it a educated guess ?
i didn't say it's not A ,B C or D :=) according the information which one do you think it's ? i repeat if he is negative then graph shift to the right side if it's positive then graph shift to the left side |dw:1444244253550:dw|
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