# Thread: transform an absolute value

1. ## transform an absolute value

The equation for this graph
is:
$\displaystyle \frac {4\mid x + 1\mid(x +3 )(x - 1)^2}{(x + 1)^4(x - 3)}$

I am trying to get rid of the absolute value of (x+1) to form a rational function. But without the absolute value, the left third of the graph is flipped over the x-axis. Is there a way to make this same graph without the absolute value?

2. Originally Posted by cubs3205
The equation for this graph
is:
$\displaystyle \frac {4\mid x + 1\mid(x +3 )(x - 1)^2}{(x + 1)^4(x - 3)}$

I am trying to get rid of the absolute value of (x+1) to form a rational function. But without the absolute value, the left third of the graph is flipped over the x-axis. Is there a way to make this same graph without the absolute value?

the $\displaystyle (x + 1)^4$ in the demominator is always positive, so you can cancel one of those with the $\displaystyle |x + 1|$. so your graph will become: $\displaystyle \frac {4(x +3 )(x - 1)^2}{(x + 1)^3(x - 3)}$

3. yes I tried that but the left side is inverted. It is sloping up rather than sloping down like in the absolute value equation.

4. Yes you are correct. There is more to it.
$\displaystyle \frac{{\left( {4\left| {x + 1} \right|\left( {x + 3} \right)\left( {x - 1} \right)^2 } \right)}}{{\left( {x + 1} \right)^4 \left( {x - 3} \right)}} =$$\displaystyle \left\{ {\begin{array}{lr} {\frac{{\left( { - 4\left( {x + 3} \right)\left( {x - 1} \right)^2 } \right)}}{{\left( {x + 1} \right)^3 \left( {x - 3} \right)}}} & {x < - 1} \\ {\frac{{\left( {4\left( {x + 3} \right)\left( {x - 1} \right)^2 } \right)}}{{\left( {x + 1} \right)^3 \left( {x - 3} \right)}}} & {x > - 1,\;x \not= 3} \\ \end{array}} \right.$