# Did I explain this right? Basic absolute value functions

1. Oct 18, 2011

### SMA_01

1. The problem statement, all variables and given/known data

I was helping my sister with her high school geometry homework, and she had to graph the absolute value function y=lx+6l using two linear equations. The way i did it was different than the way her book showed. Basically, I told her that when removing the absolute value you can get to distinct equations:

y=x+6 and -y=x+6 which becomes y=-x-6

I then told her that you can find the intersection point by setting the equations equal to each other, and solve for x. After getting the x-coordinate, you plug it back into the original equation to find the y coordinate. This will then give you the (x,y) coordinate point at which these two lines will intersect.

So at (-6,0) these lines will intersect.

I then told her to graph the two lines, making sure they intersect at this point.

To find the region on the graph that represents the absolute value function, I explained to her that y≥0. So the region on the graph that satisfied that condition represented the absolute value function.

Now, her teacher said something along the line that my method was wrong? Can someone please clarify what I did wrong? Thanks.

2. Oct 18, 2011

### Ray Vickson

Your explanation was much, much longer then needed, and is also unneccessarily complicated. An absolute value vanishes only when the thing inside the | | signs is zero, so y = 0 only at x = -6. When the thing inside | | is positive, the absolute value equals the thing itself; otherwise, the absolute value is the negative of the thing (and will, therefore, be positive---since it would be the negative of a negative thing). So, for x > -6 we have y = x+6 but for x < -6 we have y = -(x+6) = -6 - x. The graph will be like that of y = |x|, but just shifted 6 units to the left.

RGV

3. Oct 18, 2011

### SMA_01

@Ray: Thanks, I tried to explain it to her through methods she already knows, that way she won't forget it as easily. I kind of lost you at the end, though, can you please clarify that?

Also, was my method considered "wrong"?

4. Oct 18, 2011

### Ray Vickson

I can't see anything wrong with your method, but as I said, it is much more complicated than it needs to be. If the teacher says it is wrong, that just shows that either the teacher does not know the subject matter, or else your sister explained it to the teacher incorrectly or presented it wrongly, or something.

As to your being lost at the end of my message: would that be my statement that the graph of y = |x+6| is just the graph of y = |x|, but shifted 6 units to the left? What is confusing about that? The graph of y = |x| looks like the letter 'V', with vertex at x = 0 and y = 0. The graph of y = |x + 6| looks like the letter 'V'; but with vertex at x = -6 and y = 0. It really is the first graph moved 6 units to the left.

By the way: you said your method was different from that in the book. What WAS the method in the book?

RGV

5. Oct 19, 2011

### SMA_01

I understood that it shifts, and I reread it and understood the end thanks! Her book showed a method using a formula to compute the vertex, and there was another method that honestly lacked detail so I couldn't get the point. I will try to look it over again, and see how they do it. Thanks again

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