Question 255156
y=|-3x| 
<pre><font size = 4 color = "indigo"><b>
First find the VERTEX by taking what's between the absolute
value bars and putting it equal to zero:

-3x = 0

Divide both sides by -3

  x = 0

Substitute this in the original equation:

y = |-3(0)|

y = |0|

y = 0

So plot the point (0,0).  That happens to be the origin:

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5), 

line(0+.1,0,0-.1,0),  line(0,0+.1,0,0-.1),  line(0+.1,0+.1,0-.1,0-.1), 
line(0+.1,0-.1,0-.1,0+.1) )}}}

Next choose one value of x less than 0 which will give a point
on the left of the vertex, then choose another value of x greater
than 0 which will give a point on the right of the vertex:

For the point on the left of the vertex, we choose x=-1. Substitute
it in the original equation:

y = |-3x|
y = |-3(-1)|
y = |3|
y = 3

That gives you the point (-1,3) which is left of the vertex.

Next choose one value of x greater than 0 which will give a point
on the right of the vertex.

For the point on the right of the vertex, we choose x=+1. Substitute
it in the original equation:

y = |-3x|
y = |-3(+1)|
y = |-3|
y = 3

That gives you the point (1,3) which is right of the vertex. 
Plot those points:

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5), 

line(0+.1,0,0-.1,0),  line(0,0+.1,0,0-.1),  line(0+.1,0+.1,0-.1,0-.1), 
line(0+.1,0-.1,0-.1,0+.1),

line(1+.1,3,1-.1,3),  line(1,3+.1,1,3-.1),  line(1+.1,3+.1,1-.1,3-.1), 
line(1+.1,3-.1,1-.1,3+.1),

line(-1+.1,3,-1-.1,3),  line(-1,3+.1,-1,3-.1),  line(-1+.1,3+.1,-1-.1,3-.1), 
line(-1+.1,3-.1,-1-.1,3+.1)  )}}}

Draw the v-shaped graph:

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5,-6,abs(-3x)), 

line(0+.1,0,0-.1,0),  line(0,0+.1,0,0-.1),  line(0+.1,0+.1,0-.1,0-.1), 
line(0+.1,0-.1,0-.1,0+.1),

line(1+.1,3,1-.1,3),  line(1,3+.1,1,3-.1),  line(1+.1,3+.1,1-.1,3-.1), 
line(1+.1,3-.1,1-.1,3+.1),

line(-1+.1,3,-1-.1,3),  line(-1,3+.1,-1,3-.1),  line(-1+.1,3+.1,-1-.1,3-.1), 
line(-1+.1,3-.1,-1-.1,3+.1)  )}}}





40. y=-|x-3| 

First find the VERTEX by taking what's between the absolute
value bars and putting it equal to zero:

x-3 = 0

Add +3 to both sides:

  x = 3

Substitute this in the original equation:

y = -|(3)-3|

y = -|0|

y = 0

So plot the point (3,0)

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5), 

line(3+.1,0,3-.1,0),  line(3,0+.1,3,0-.1),  line(3+.1,0+.1,3-.1,0-.1), 
line(3+.1,0-.1,3-.1,0+.1) )}}}

Next choose one value of x less than 3 which will give a point
on the left of the vertex, then choose another value of x greater
than 3 which will give a point on the right of the vertex:

For the point on the left of the vertex, we choose x=2. Substitute
it in the original equation:

y = -|x-3|
y = -|(2)-3|
y = -|2-3|
y = -|-1|
y = -(1)
y= -1

That gives you the point (2,-1) which is left of the vertex.

Next choose one value of x greater than 3 which will give a point
on the right of the vertex.

For the point on the right of the vertex, we choose x=4. Substitute
it in the original equation:

y = -|x-3|
y = -|(4)=3|
y = -|4-3|
y = -|1|
y = -1

That gives you the point (4,-1) which is right of the vertex. 
Plot those points:

{{{drawing(400,400,-3,7,-5,5, graph(400,400,-3,7,-5,5), 

line(3+.1,0,3-.1,0),  line(3,0+.1,3,0-.1),  line(3+.1,0+.1,3-.1,0-.1), 
line(3+.1,0-.1,3-.1,0+.1),

line(4+.1,-1,4-.1,-1),  line(4,-1+.1,4,-1-.1),  line(4+.1,-1+.1,4-.1,-1-.1), 
line(4+.1,-1-.1,4-.1,-1+.1),

line(2+.1,-1,2-.1,-1),  line(2,-1+.1,2,-1-.1),  line(2+.1,-1+.1,2-.1,-1-.1), 
line(2+.1,-1-.1,2-.1,-1+.1)  )}}}

Draw the v-shaped graph:

{{{drawing(400,400,-3,7,-5,5, graph(400,400,-3,7,-5,5,-abs(x-3)), 

line(3+.1,0,3-.1,0),  line(3,0+.1,3,0-.1),  line(3+.1,0+.1,3-.1,0-.1), 
line(3+.1,0-.1,3-.1,0+.1),

line(4+.1,-1,4-.1,-1),  line(4,-1+.1,4,-1-.1),  line(4+.1,-1+.1,4-.1,-1-.1), 
line(4+.1,-1-.1,4-.1,-1+.1),

line(2+.1,-1,2-.1,-1),  line(2,-1+.1,2,-1-.1),  line(2+.1,-1+.1,2-.1,-1-.1), 
line(2+.1,-1-.1,2-.1,-1+.1)  )}}}

Edwin</pre>