Explain how the graph of y=|x+1|-2 will differ from the graph of y=|x|.

with that template in mind,

so is really just |x| but shifted some,

C/B = 1/1 = +1, horizontally to the right by 1 unit.

D = -2, vertically downwards by 2 units.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

y = i/2 xtnop + (3i/2 tnop+ 1)

step-by-step explanation:

the picture is the step by step explanation.