Absolute value graph: y = - l x - 2 l

A

AGlas9837

Junior Member
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Jan 23, 2008
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The instructions for this problem are to sketch the graph of the equation and label the intercepts, then use a graphing utility to verify results.

y=-lx-2l - I'm using l's for the absolute value symbols. For the x-intercept, I got:

0=-1lx-2l
0=-2
(x-intercept = (-2,0)

For the y-intercept, I got:

y=-l0-2l
y-intercept = (0,-2)

When I used the points to graph, the "point" (?vertex) of the graph was at x = -2. When I plugged in the equations to graph, the point of the graph was located at x = 2. Are my intercepts correct and how do I graph?
 
AGlas9837 said:
y=-lx-2l... For the x-intercept, I got:

0=-1lx-2l
0=-2
Click to expand...
What happened to your variable? Since zero does not, in fact, ever equal negative two, what do you mean by "0 = -2"? :shock:

AGlas9837 said:
how do I graph?
Click to expand...
Graph in the usual manner: Pick some x-values, plug them into the equation, solve for the corresponding y-values, draw the dots, and then sketch the graph. :wink:

Eliz.
 
hopefully, you have covered transformations of functions in class.

y = -|x - 2| is the graph of y = |x| that has undergone two transformations ... a horizontal translation to the right 2 units, and a reflection over the x-axis,
 
y = -|x - 2|

y = -|0 - 2|
y = -|-2|
y = -2 when x = 0
(0,-2)

when y = 0
0 = -|x - 2|
what minus 2 = 0 ? forget about the absolute at this point and the sign
the only solution for x would be 2 right?
so when y = 0 x = 2
0 = -|2 -2|
0 = -|0|
since 0 is the same number whether negative or positive.
(2,0)
 
Let's back up a bit and be very, very careful and deliberate.

y = -|x-2|

The difficulty in graphing is the absolute value. Let's deal with them and not ever "forget about" them.

The fundamental principle is this

If x >= 0, then |x| = x. Example: |3| = 3.

Further, if x < 0, then |x| = -x. Example: |-4| = -(-4) = 4

back to the graph.

y = -|x-2|

Our task is to split things up so they make sense individually. With the fundamental principle in mind...

x-2 >= 0 where x >= 2 so that |x-2| = (x-2)
x-2 < 0 where x < 2 so that |x-2| = -(x-2) = (-x+2)

Now we have the graph in two pieces

y = -(x-2) where x >= 2 AND y = -(-x+2) where x < 2

You should see the direct application of the fundamental principle.

Simplify a little

y = -x+2 where x >= 2 AND y = x-2 where x < 2

Simply graph these two lines and cross out the parts that don't apply.

1) Graph y = -x+2 and cross out everything where x < 2
2) Graph y = x-2 and cross out everything where x >= 2

No short cuts. Think about every step.
 
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