\right) & =-|x|\left(-|x|-a\right)<br />\\ & =-|x|\times-\left(|x|+a\right)<br />\\ & =|x|\left(|x|+a\right)<br />\end{alignedat}<br />)
Now to work out how to graph our result. Normally if we have
|)
, then we'd plot the curve of
)
and then flip everything that is below the
axis in the
axis. We can do something similar with
)
, except that all the inputs for negative
will now have the same output for the corresponding positive value of
. That is you would draw the original curve, then take the curve that is on the right side of the
axis, copy it and flip it in the
axis.
So back to our problem, we can recognise that if we were dealing with say
)
then,
=x(x+a))
. So firstly we should sketch
)
https://www.desmos.com/calculator/gq69kjxgexNow we take everything that is on the right hand side of of the
axis,
https://www.desmos.com/calculator/5bamx2wuauAnd copy it, flipping it in the
axis.
https://www.desmos.com/calculator/teo9gw6ryqWhich we can see is the same as what we were looking for originally.
https://www.desmos.com/calculator/ebcda7ctrhHope that makes sense and hope that helps!
Edit: You could probably split it up into a hybrid function, but it'd be a bit more tedious. For me I find that applying the transformations to the curve when working with modulus functions works best, but thats just me. It may or may not be the best method for you.
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