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how to sketch: $-e^{|-x-1|} + 2$

Can someone clarify:

$|f(x)|:$ we draw $f(x)$ and then reflect the ($-y$ parts) in the $x$-axis

$f|(x)|:$ we draw $f(x)$ and then reflect the ($-x$ parts) in the $y$-axis (symmetry on left and right hand side), can someone correct me here!

The Steps to sketch the above equation:

  1. The original equation is e^x which then becomes e^|x| which shows to have undergone f|(x)|

  2. We sketch -e^(-x-1) without applying the absolute value

  3. Then we apply the absolute value, by reflecting along the turning point

  4. Then we shift the graph up 2 units

so basically:

  1. draw the e^x graph
  2. apply the reflections/dilations/horizontal translations
  3. apply f(|x|)
  4. apply the vertical transformation

can someone correct me here.

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That would be graph for $e^{-|x|-1}+2$.

What you want to do is check when is $|-x-1|$ = $1+x$ or $-1-x$ and draw those $2$ graphs separately.

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