Use graphing to solve: a. |x+2|=6 b.|x+2|>6

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Use graphing to solve: a. |x+2|=6 b.|x+2|>6

Mathematics
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graph the y = |x + 2| and graph y = 6 |dw:1328675607779:dw| find the points of intersection (4, 6) and (-8, 6) so read off the graph... when is |x + 2| . 6.... when x > 4 and -8
but how do you know where the arrow goes?
pick a point.... say (1, 1) sub x= 1 into |x + 2|>6 then | 1+2| > 6 T or F... its false so (1,1) isn't in the region... choose a point test it and see if its in or out of the region...

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Other answers:

uh i dont get it
look at the graph... when is the graph of |x +2| above the line y = 6..... and describe the x values.... then graph is above the line y = 6 when x > 4 and also when x < -8
but how do you know how far the arrow will stretch? why (4,6) and (-8,6)?
well the domain of |x + 2| is all real x.... an inequality can have infinite solutions.
yeah but why (4,6)? why not (6,4) or something?
|x + 2|= 6 is really asking where the lines y=|x +2| and y=6 intersect... that was the 1st task in your question, solve graphically.... sou graoh both and read off the points intersections... then its is asking.. when is |x + 2| greater then 6... or really when is it greater than the line y = 6

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