Abs Graphing (1 Viewer)

coyazayo · Jan 22, 2016

Hi to everyone reading this thread. I was wondering what the graphss would look like or the following:

1. y=(|x|)^3+1
2. y=|x^3+1|
3.|y|=x^3+1

InteGrand · Jan 22, 2016

coyazayo said:
Hi to everyone reading this thread. I was wondering what the graphss would look like or the following:

1. y=(|x|)^3+1
2. y=|x^3+1|
3.|y|=x^3+1

$\noindent 1. For $x\geq 0$, it just looks like the graph of $y=x^3+1$. For $x<0$, reflect the $x\geq 0$ part of the graph about the $y$-axis. So it looks like this:$

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

$\noindent 2. Sketch the graph of $y=x^3 + 1$ first. Then for any part of this graph that's below the $x$-axis, flip it about the $x$-axis. The parts above the $x$-axis stay as they are. So it looks like this $\Big{(}$the red graph is that of $y=x^3 + 1$. The blue part combined with the red part above the $x$-axis is the graph of $y=\left |x^3 + 1\right|$$\Big{)}$:$

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

$\noindent 3. This is the graph of $y=\pm \left(x^3 + 1\right), \,x^3+1\geq 0\Longleftrightarrow x\geq -1$, because $|y| = x^3 + 1$ if and only if $y=\pm \left(x^3 + 1\right) $ and $x^3 + 1\geq 0$. So sketch the graph of $y=x^3 + 1$, and also the graph of $y=-\left(x^3 + 1\right) = -x^3 -1$. These two graphs together are the graph of the curve you want, but include and only include the parts of the graphs where $x\geq -1$.$

braintic · Jan 23, 2016

coyazayo said:
Hi to everyone reading this thread. I was wondering what the graphss would look like or the following:

1. y=(|x|)^3+1
2. y=|x^3+1|
3.|y|=x^3+1

Just to make sure you are aware of an issue that wasn't brought out by the other questions, graph |y| = x^3 - 1

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Abs Graphing (1 Viewer)

coyazayo

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InteGrand

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braintic

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