Thread: Absolute value graphs

1. y = ABS(x/2 + 5) - 3
i got a v shape, need confirmation on this one

2. ABS(2x - 3) + ABS(x + 3) - ABS(x + 1) - 3

i dont know how to graph this. I need to show work

3. Write y from #2 as a split rule without any absolute value bars.

I dont even know what split rule is????


Any help is appreaciated

Ok, i think i got number 2
I found the turning points and connected them but i still think i need to find the equations for the connected lines for problem 2. How do i do that?

Originally Posted by mathprob

1. y = ABS(x/2 + 5) - 3
i got a v shape, need confirmation on this one

Since we cannot see your work, we cannot confirm your graph. Sorry!

However, you should be able to confirm the graph in your graphing calculator or some other software.

Originally Posted by mathprob

2. ABS(2x - 3) + ABS(x + 3) - ABS(x + 1) - 3

i dont know how to graph this. I need to show work

Since there is no "y=" or "f(x)=" or something similar, it is difficult to be sure how to proceed. But the process is (probably) the usual: pick x-values, plug them into the rule, simplify to find the corresponding y-values, draw the dots, etc, etc. Try here for specifics.

Note: Because you have three absolute-value expressions here, you'll likely want to pay particular attention to the "interesting" points, being where the arguments change sign. For instance, |2x - 3| is negative for x < 3/2 and positive for x > 3/2. So be sure to include x = 3/2 in your plot-points.

Originally Posted by mathprob

3. Write y from #2 as a split rule without any absolute value bars.

I dont even know what split rule is?

If your book didn't define this, then your instructor must be the source, so ask there for clarification.

Thank you!