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:
-
The original equation is e^x which then becomes e^|x| which shows to have undergone f|(x)|
-
We sketch -e^(-x-1) without applying the absolute value
-
Then we apply the absolute value, by reflecting along the turning point
-
Then we shift the graph up 2 units
so basically:
- draw the e^x graph
- apply the reflections/dilations/horizontal translations
- apply f(|x|)
- apply the vertical transformation
can someone correct me here.
Best Answer
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.