How Would I Graph This Exponential Function?

$f(x) = \frac{-3}{2^{(x+2)}} - 1$,

How I do it: I know that it is basically $-3\times ({\frac{ 1}{2}})^{x+2} - 1$, which is then $-3 \times 2^{-(x+2)} - 1$, my base function is $2^{-x}$, and thus my mapping equation ends up being $(-x + 2, -3y - 1)$, and after plugging in points such as $(-2, 4), (-1, 2), (0, 1), (1, \frac{1}{2})$, etc, I end up with my graph but it looks completely different than the one I get in Wolframalpha - well, similar, but it is starting from the top left and it gradually proceeds down. What am I doing wrong?


  1. First graph $\frac{1}{2^x}$. This is easy.
  2. Next, slide the graph over 2 units to the left. This results in $\frac{1}{2^{x+2}}$.
  3. Next, scale the graph by 3, this results in $\frac{3}{2^{x+2}}$.
  4. Now, flip the graph, yielding $-\frac{3}{2^{x+2}}$.
  5. Finally, slide the graph down by $1$, to obtain $-\frac{3}{2^{x+2}}-1$.

To wit:

  1. $x=2 \implies y = \frac14$. $x=3 \implies y=\frac18$.
  2. Slide left by 2: $x=0 \implies y = \frac14$. $x=1 \implies y=\frac18$.
  3. Scale by 3: $x=0 \implies y = \frac34$. $x=1 \implies y=\frac38$.
  4. Flip it: $x=0 \implies y = -\frac34$. $x=1 \implies y=-\frac38$.
  5. Slide down by 1: $x=0 \implies y = -\frac74$. $x=1\implies y=-\frac{11}{8}$

After working this out, it appears that maybe you had an extra negative sign in your computation of $x$. Remember, $2^{-x} = \frac{1}{2^x}$, so if you perform $x \mapsto x+2$, then make sure you either compute $x+2$ and plug it into $2^{-x}$, or compute $-x-2$ and plug it into $2^x$.

  • $\begingroup$ Oh, okay, so having 2^(x+2), by 'flipping', would I have -x-2, or -x + 2? $\endgroup$ – user164403 May 19 '15 at 23:20
  • $\begingroup$ Neither. "Flipping" is done to the $y$ values. Note that I started my algorithm by computing $\frac{1}{2^x}$, not $2^x$. Computing powers of $\frac12$ is easy enough that we can start there. $\endgroup$ – Emily May 19 '15 at 23:22
  • $\begingroup$ Note that this is an exercise in the following generalization: $f(x+a)$ slides the graph of $f(x)$ by $a$ units to the left; $f(cx)$ makes a graph wider by a factor of $c$; $f(x)+b$ slides the graph $b$ units up; $df(x)$ makes the graph taller by a factor of $d$. In each case, negative values of $a,b,c,d$ do the opposite; i.e. $f(x-a)$ slides the graph to the right. $\endgroup$ – Emily May 19 '15 at 23:27

My goodness... so much work when we have computer plotting right at hand?! If a question poser asked how to multiply two 10-digit numbers would we explain how to do it by hand? enter image description here

Your points are in red, which is fit perfectly by $y = 2^{-x}$ (in orange).

  • $\begingroup$ This does not provide the user with an answer to the question "what am I doing wrong." The OP explicitly mentions that they used W|A to validate a solution; hence this post adds no substance. Maple exists, why learn calculus? $\endgroup$ – Emily May 19 '15 at 23:24
  • $\begingroup$ One cannot answer the student's question adequately without see his/her graph.... or at least I cannot. How does his/her graph differ from the one given by Wolfram|Alpha? Might he/she have typed the equation into Wolfram|Alpha incorrectly? $\endgroup$ – David G. Stork May 19 '15 at 23:26
  • $\begingroup$ You have a computer at your disposal. Why not plot the points they gave in their post to approximate? $\endgroup$ – Emily May 19 '15 at 23:28
  • $\begingroup$ I don't understand where I made the sign error. My mapping equation ends up being (-x + 2, -3y - 1), and one of my first points is (0,1). I express 1/2^(x+2) as 2^-x-2, so doesn't seem like I did anything wrong... I don't get where I'm going wrong. $\endgroup$ – user164403 May 19 '15 at 23:46

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