## Friday, March 5, 2010

### Absolute value

I spent a fair bit of time thinking about absolute value problems in the form |x+a| - |x-b| = c. Many students were struggling with visualising what these functions actually look like. What was happening when we try and solve them?

For example:
|x+5| - |x - 2| = 6
How could I display this equation graphically to give students an understanding of the underlying algebra to solve it?

I tried graphing y = |x+5| - |x - 2| and y=6 to find the intersection but was unsatisfied with the result as y = |x+5| - |x - 2| is not something easily tied to the absolute value concept or 'v' shaped absolute value graphs.
I was eventually satisfied with graphing y= |x+5| and y = |x-2| and then examining each part of the graph until I found a section of the graph that was 6 units apart.

For those wondering how to put it into a graphics calculator while exploring the concept

Go Menu -> Graph & Tab
Edit -> Clear All -> ok
at "Y1:" ->Softkeyboard->mth tab->select 'x'->type "x+5)" (it will change from abs(x+5) to |x+5|)
at "Y2:"->select 'x'->type "x-2)"
ensure that the boxes next to "Y1" and "Y2" are ticked

Now the temptation is to assume the answer is the intersection point.

but if we look at the equation |x+5| - |x - 2| = 6, it is asking "for what value of x is the value of |x+5| (the dotted line) subtract the value of |x-2| (the solid line) equal to 6". When is the gap between the two functions +6.

We can ignore values of x<= -5 as y=|x+5| is below y=|x-2| and the subtraction will only give negative values (we are looking for a gap of +6 which is a positive value).

We can also ignore values up to the intersection point as this also will only result in negative values.
The next place I looked is at x>=2 as the gap is constant and positive after this point (both functions have the same gradient).
at x=2, |x+5| is equal to 7 and |x-2| is equal to 0. |2+5| - |2-2| = 7. We can ignore values where x>=2 as the answer is not +6.

In fact the only possible solution has to lie between the intersection point (x~-1.5) and 2 and is probably closer to 2.
For y=|x+5| all values are positive between -1.5 < x < 2
For y=|x-2| all values are negative between -1.5 < x < 2
To ensure positive values for x-2 in the range -1.5< x < 2 we need to take the negative of (x-2) when solving the equation |x+5| - |x - 2| = 6.

x+5 - (-(x-2)) = 6
2x+3 = 6
x=1.5

|x+5| - |x - 2| = 6
Let x=1.5
|1.5+5| - |1.5-2| = 6
LHS =| 6.5| -| -.5|
= 6.5 - 0.5
= 6.0
= RHS

Viola.

It would also be interesting to explore |x+a| + |x-b| = c,  |x+a| = |x-b| and -|x+a| = c in a similar way.

Here is a link to other CAS calculator posts.