In the equation:
#y=5abs(3x-4)#, the absolute value part of the equation (the #abs(3x-4)# part) will never return anything less than 0. And so the first thing we can do is to see what value of #x# will have the absolute value term return a 0:
#3x-4=0 => x=4/3#
And so for values of #x#, both increasing and decreasing from #x=4/3#, the absolute value term will increase. This is the #x# value of the vertex. The #y# value is:
#y=5abs(3(4/3)-4)=5xx0=0#
And so the vertex is at #(4/3,0)#
Ok - so for values of #x>=4/3#, we'll take the positive aspect of the absolute value term and distribute through:
#x>=4/3, y=5(3x-4)=15x-20#
And for #x<=4/3#, we take the negative aspect of the absolute value term:
#x<=4/3, y=5(-1(3x-4))=-15x+20#
Let's compare the graphs to make sure we've got this correct. The top one will be the absolute value graph and the bottom will have both line graphs (keep in mind the negative #y# values in the second graph are simply continuations of the line equations beyond the stated boundaries of #x# for each line):
graph{5abs(3x-4)[-5,10,-5,20]}
graph{(y-(15x-20))(y-(-15x+20))=0 [-5,10,-5,20]}
