# Graph the function $f(x)=|e^{-x+2}-1|$ without using graphing calculator.

Graph the function $f(x)=|e^{-x+2}-1|$ without using graphing calculator.

I tried so many times to solve it but $i$ was really confused because of the absolute value??

SO, Could you please tell the procedures that I should follow in graphing functions containing Absolute value like the one above

• If you can get to the point of considering the effect of the absolute value, you are nearly done. In general, $|f(x)|$ will "reflect" negative portions of the graph of $f$ vertically over the $x$-axis, effectively making any negative $y$-values positive. – Kaj Hansen May 13 '14 at 3:33

## 3 Answers

Hints:

Since $\;e^t\ge 1\iff t\ge0\;$ , we have that

$$e^{-x+2}-1\ge0\iff x\le 2$$ so

$$\left|e^{-x+2}-1\right|=\begin{cases}e^{-x+2}-1&,\;\;x\le 2\\{}\\1-e^{-x+2}&,\;\;x>2\end{cases}$$

procedure

a) Hm, $-x+2=(-(x-2))$ means a shift by 2 and a reflection

b) So now I have an idea what $e^{-x+2}$ looks like

c) $-1$ ok, move it down one unit - got $e^{-x+2}-1$

d) absolute value...have to know where $f(x)=e^{-x+2}-1<0$ and draw the graph at these points above the $x$-axis

Try to break the problem: 1. plot exp(-x) It would be like this: 1. multiply exp(-x) and exp(2) and plot: 1. subtract 1 from this on every point of the graph. Then, for -ve x, not much difference would be visible, as f(x) is very high valued. But for +ve x, as we go towards right, f(x) becomes -ve (which was not happening earlier). 1. now take absolute of this. For absolute: $$abs(x)=\begin{cases}-x&,\;\;x<0\\{}\\x&,\;\;x>0\end{cases}$$ Hence, only the f(x) which was -ve in the above graph will go +ve. everything else remains same: this will be the final graph.