# Thread: Graphing absolute values (pictures)

1. ## Graphing absolute values (pictures)

Two questions
Is the first one right, I did it with one teacher using a t chart and that's the solution I got.

First, Can you solve for y and get an accurate graph on a graphing calculator(where there are no points plotted that should not be)

I can see some working on the calculator but in this example and one other there are points which are graphed that do not fulfill the initial equation.

for example (2,1)

People tell me not to always use t charts but with examples like this I question not using t charts.

2. Both graphs have some points in common. The second graph appears to be the first graph, but with a few lines extended.

I believe (2, -1) is a point on your second graph. Does plugging these values in satisfy the equation?

3. Originally Posted by wytiaz
Both graphs have some points in common. The second graph appears to be the first graph, but with a few lines extended.

I believe (2, -1) is a point on your second graph. Does plugging these values in satisfy the equation?
no, thats whats confusing me.
What I want to know is if you are given the equation |x|-|y|=1
how you would convert that to give you an equation you can put into the tI-83 calculator. The way he did it it shows points that dont satisfy the initial equation.

4. Originally Posted by brentwoodbc
What I want to know is if you are given the equation |x|-|y|=1 how you would convert that to give you an equation you can put into the tI-83 calculator.
Trying to fit this into your calculator may be the longer way of doing this! Instead, try working with the algebra of absolute values and absolute-value functions.

Consider cases:

i) If $\displaystyle y\, \geq\, 0$, then $\displaystyle |y|\, =\, y$ and you have $\displaystyle |x|\, -\, y\, =\, 1$, or $\displaystyle y\, =\, |x|\, -\, 1$.

a) If $\displaystyle x\, \geq\, 0$, then $\displaystyle |x|\, =\, x$ and the equation in (i) becomes $\displaystyle y\, =\, x\, -\, 1$.

b) If $\displaystyle x\, <\, 0$, then $\displaystyle |x|\, =\, -x$ (since you'd have to change the sign on the negative value of x to get the positive absolute value), and the equation in (i) becomes $\displaystyle y\, =\, -x\, -\, 1$.

ii) If $\displaystyle y\, <\, 0$, then $\displaystyle |y|\, =\, -y$ (since you'd have to change the sign on the negative value of y to get the positive absolute value), and the equation becomes $\displaystyle |x|\, -\, (-y)\, =\, 1$. Then $\displaystyle |x|\, +\, y\, =\, 1$, so $\displaystyle y\, =\, 1\, -\, |x|$.

a) If $\displaystyle x\, \geq\, 0$, then $\displaystyle |x|\, =\, x$ and the equation in (ii) becomes $\displaystyle y\, =\, 1\, -\, x$.

b) If $\displaystyle x\, <\, 0$, then $\displaystyle |x|\, =\, -x$ (since you'd have to change the sign on the negative value of x to get the positive absolute value), and the equation in (ii) becomes $\displaystyle y\, =\, 1\, -\, (-x)$, or $\displaystyle y\, =\, 1\, +\, x$.

If you solve the pairs of cases:

$\displaystyle x\, <\, 0:\, y\, =\, -x\, -\, 1\, \mbox{ and }\, y\, =\, 1\, +\, x$

$\displaystyle x\, \geq\, 0:\, y\, =\, x\, -\, 1\, \mbox{ and }\, y\, =\, 1\, -\, x$

...you get $\displaystyle -x\, -\, 1\, =\, 1\, +\, x$ for the first pair, which simplifies as $\displaystyle -2\, =\, 2x$, so $\displaystyle x\, =\, -1$, and you get $\displaystyle x\, -\, 1\, =\, 1\, -\, x$ for the second pair, which simplifies as $\displaystyle 2x\, =\, 2$, so $\displaystyle x\, =\, 1$.

Hope that helps!

5. Originally Posted by brentwoodbc
no, thats whats confusing me.
What I want to know is if you are given the equation |x|-|y|=1
how you would convert that to give you an equation you can put into the tI-83 calculator. The way he did it it shows points that dont satisfy the initial equation.
You have to transform the equation a little bit using the definition of the absolute value of y:

$\displaystyle |x|-|y|=1~\implies~|y|=|x|-1$

$\displaystyle |y| = \left\{\begin{array}{l}1-|x|,\ if\ y\geq 0 \\|x|-1,\ if\ y<0\end{array} \right.$

These two terms have to written in the Y= editor of the TI83 together with the domain. Write at

$\displaystyle y1=\text{(abs(x)-1)(-1}\leq \text{x)(x} \leq\text{1)}$

$\displaystyle y2=\text{(1-abs(x))(-1}\leq \text{x)(x} \leq\text{1)}$

You find abs( in the catalogue. You only can use the $\displaystyle \leq$-relation to determine the domain.

I've attached a screen-shot of the input-screen, the window settings and the graph.

6. Originally Posted by brentwoodbc
no, thats whats confusing me.
What I want to know is if you are given the equation |x|-|y|=1
how you would convert that to give you an equation you can put into the tI-83 calculator. The way he did it it shows points that dont satisfy the initial equation.
If $\displaystyle y\ge 0$ then $\displaystyle y=|x|-1$. If $\displaystyle y< 0$ then $\displaystyle |y|=-y=|x|-1\Rightarrow y=1-|x|$.
So put two functions into your TI-83: $\displaystyle Y_1=\text{abs}(X)-1$ and $\displaystyle Y_2=1-\text{abs}(X)$.

EDIT: earboth's solution is better!

7. Thanks guys so much, esp. earboth. I appreciate it alot, this stuff has been driving me nuts.

You explained it much better then 3 of my math teachers combined. haha