How do you graph #h(x)= -5/3 absx + 6#?

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Answer 1 · 2018-08-10T03:41:15

See explanation and #angle#-shaped graph.

Here,

#abs x = 3/5( 6 - h ) >=0 rArr h <= 6#. Use, y for h.

Socratic inverted V graph is immediate.
graph{(abs (x) -0.6(6-y))(y-6+0x)=0[-16 16 -8 8]}

The vertex is ( 0, 6 ). You can see the graph, with #h <= 6#.

If you change #abs x to x#, the graph would be as shown below.
graph{(x - 0.6(6-y))(x+0.6(6-y))=0[-10 20 0 15]}

For this, the equation is #(x-0.6(6-h)(x+0.6(6-h)) = 0#,

for the whole patr of straight lines..