#color(blue)("Explaining what each part of the equation does.")#

#color(blue)("ie the transformation from basic form")#

This is transforming the graph of #y=x# which is the general shape of the sloping line type /

Using the absolute format #y=|x| # changes the general shape to V where the vertex ( point ) is at the x-axis. So for #ul("this case")#:

Vertex #->(x,y)=(0,0)# and y is always positive

Including the + 3 #->y=|x+3|# 'slides' the graph to the left by 3.

Adding 'slides' to the left whilst subtracting 'slides' it to the right.

Why is this? Consider the graph of #y=|x|#. Suppose we had the x-value of 0. Move to the right by 3. Draw a faint line upwards and note the point on the graph. Now plot a point at that y-value where#x=0#. You have 'shifted' to the left by 3.

The constant of -1 drops the whole thing by 1.

So for #y=|x+3|-1# the value if #y# can now be negative but no less than -1.

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#color(blue)("Determine the vertex")#

Using the above explanation the Vertex #->(x,y)=(-3,-1)#

Or you cam calculate it by setting #y=-1#

#-1=x+3-1#

#x=-1-3+1=-3#

'......................................................................................

#color(blue)("Determine y-intercept")#

Set #x=0#

#y_("intercept") = |0+3|-1 = 2#

'.......................................................................................

#color(blue)("Determine x-intercept")#

Set #y=0#

#0=|x+3|-1#

Consider just one side of the V

Set #y=x+3-1" "->" "y=0=x+2#

In this case #x=-2#

Compare to the vertex #->(x,y)=(-3,-1)#

The point #x=-2# is to the right of #x=-3# by 1.

So the other point must be to the left of #x=-3# by 1 as well

Thus #x_("intercept") -> x=-4 and x=-2#