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Math Help - transform an absolute value

  1. #1
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    transform an absolute value

    The equation for this graph
    is:
    <br />
\frac {4\mid x + 1\mid(x +3 )(x - 1)^2}{(x + 1)^4(x - 3)}<br />

    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?

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cubs3205 View Post
    The equation for this graph
    is:
    <br />
\frac {4\mid x + 1\mid(x +3 )(x - 1)^2}{(x + 1)^4(x - 3)}<br />

    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  (x + 1)^4 in the demominator is always positive, so you can cancel one of those with the |x + 1|. so your graph will become:  \frac {4(x +3 )(x - 1)^2}{(x + 1)^3(x - 3)}
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  3. #3
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    yes I tried that but the left side is inverted. It is sloping up rather than sloping down like in the absolute value equation.
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  4. #4
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    Yes you are correct. There is more to it.
    \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)}} = \left\{ {\begin{array}{lr}<br />
   {\frac{{\left( { - 4\left( {x + 3} \right)\left( {x - 1} \right)^2 } \right)}}{{\left( {x + 1} \right)^3 \left( {x - 3} \right)}}} & {x <  - 1}  \\<br />
   {\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}  \\<br />
\end{array}} \right.
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