Question 976010
{{{y=(3.4)abs((0.7)(x-4))+1}}}


Critical x value will be for the absolute value part to be 0.  That happens at x=4.  The graph has two branches.  For {{{x<4}}}, and for {{{x>=4}}}



For {{{x>=4}}}, expression inside abs val is positive or zero.
{{{y=3.4((0.7)(x-4))+1}}}
{{{y=3.4(0.7x-2.8)+1}}}
{{{y=2.38x-9.52+1}}}
{{{y=2.38x-8.52}}}-----and you ONLY want the part where {{{x>=4}}}; NOTHING to the left.


For {{{x<4}}}, expression inside the abs val is NEGATIVE.
{{{y=3.4((-1)(0.7)(x-4))+1}}}
{{{y=3.4(-.7x+2.8)+1}}}
{{{y=-2.38x+9.52+1}}}
{{{y=-2.38x+10.52}}}------and you ONLY want the part to the left of {{{x=4}}}; nothing to the right.


{{{graph(300,300,-4,10,-2,12,(3.4)abs((0.7)(x-4))+1)}}}


DISCUSSION OF TRANSFORMATION
The equation and graph fundamentally starts from {{{y=abs(x)}}}.  This would be a "V" shaped figure, with the sharp corner at the origin, and both parts are above the x-axis, and therefore, y is non-negative everywhere.  Shift this 4 units to the right, and 1 unit up, and you are at {{{y=abs(x-4)+1}}}.  Same shape, just shifted so the vertex is now at (4,1), the sharp point of the "V".
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The factor,  0.7, squashes the graph vertically; and then the factor  3.4, stretches the graph vertically, MORE SO than the factor of the 0.7 did.  This combination of factors only affects the slope or slopes of each branch.  The combined factor for this is  {{{3.4*0.7=2.38}}}; overall, this stretches the graph of {{{y=abs(x-4)}}} vertically; but still same vertex or sharp point at  (4,1).  The factor is not arranged to affect the "1".