Tag Archives: translation

Exponentials Don’t Stretch

OK, this post’s title is only half true, but transforming exponentials can lead to counter-intuitive results.  This post shares a cool transformations activity using dynamic graphing software–a perfect set-up for a mind-bending algebra or precalculus student lesson in the coming year.  I use Desmos in this post, but this can be reproduced on any graphing software with sliders.


You can vertically stretch any exponential function as much as you want, and the shape of the curve will never change!

But that doesn’t make any sense.  Doesn’t stretching a curve by definition change its curvature?

The answer is no.  Not when exponentials are vertically stretched.  It is an inevitable result from the multiplication of common bases implies add exponents property:

b^a * b^c = b^{a+c}

I set up a Desmos page to explore this property dynamically (shown below).  The base of the exponential doesn’t matter; I pre-set the base of the parent function (line 1) to 2 (in line 2), but feel free to change it.


From its form, the line 3 orange graph is a vertical stretch of the parent function; you can vary the stretch factor with the line 4 slider.  Likewise, the line 5 black graph is a horizontal translation of the parent, and the translation is controlled by the line 6 slider.  That’s all you need!

Let’s say I wanted to quadruple the height of my function, so I move the a slider to 4.  Now play with the h slider in line 6 to see if you can achieve the same results with a horizontal translation.  By the time you change h to -2, the horizontal translation aligns perfectly with the vertical stretch.  That’s a pretty strange result if you think about it.


Of course it has to be true because y = 2^{x-(-2)} = 2^x*2^2 = 4*2^x.  Try any positive stretch you like, and you will always be able to find some horizontal translation that gives you the exact same result.

Likewise, you can horizontally slide any exponential function (growth or decay) as much as you like, and there is a single vertical stretch that will produce the same results.

The implications of this are pretty deep.  Because the result of any horizontal translation of any function is a graph congruent to the initial function, AND because every vertical stretch is equivalent to a horizontal translation, then vertically stretching any exponential function produces a graph congruent to the unstretched parent curve.  That is, any vertical stretch of any exponential will never change its curvature!  Graphs make it easier to see and explore this, but it takes algebra to (hopefully) understand this cool exponential property.


My students inevitably ask if the same is true for horizontal stretches and vertical slides of exponentials.  I encourage them to play with the algebra or create another graph to investigate.  Eventually, they discover that horizontal stretches do bend exponentials (actually changing base, i.e., the growth rate), making it impossible for any translation of the parent to be congruent with the result.


But if a property is true for a function, then the inverse of the property generally should be true for the inverse of the function.  In this case, that means the transformation property that did not work for exponentials does work for logarithms!  That is,

Any horizontal stretch of any logarithmic function is congruent to some vertical translation of the original function.  But for logarithms, vertical stretches do morph the curve into a different shape.  Here’s a Desmos page demonstrating the log property.


The sum property of logarithms proves the existence of this equally strange property:

log(A) + log(x) = log(A*x)


Hopefully the unexpected transformational congruences will spark some nice discussions, while the graphical/algebraic equivalences will reinforce the importance of understanding mathematics more than one way.

Enjoy the strange transformational world of exponential and log functions!

Trig Identities with a Purpose

Yesterday, I was thinking about some changes I could introduce to a unit on polar functions.  Realizing that almost all of the polar functions traditionally explored in precalculus courses have graphs that are complete over the interval 0\le\theta\le 2\pi, I wondered if there were any interesting curves that took more than 2\pi units to graph.

My first attempt was r=cos\left(\frac{\theta}{2}\right) which produced something like a merged double limaçon with loops over its 4\pi period.

Trying for more of the same, I graphed r=cos\left(\frac{\theta}{3}\right) guessing (without really thinking about it) that I’d get more loops.  I didn’t get what I expected at all.

Wow!  That looks exactly like the image of a standard limaçon with a loop under a translation left of 0.5 units.

Further exploration confirms that r=cos\left(\frac{\theta}{3}\right) completes its graph in 3\pi units while r=\frac{1}{2}+cos\left(\theta\right) requires 2\pi units.

As you know, in mathematics, it is never enough to claim things look the same; proof is required.  The acute challenge in this case is that two polar curves (based on angle rotations) appear to be separated by a horizontal translation (a rectangular displacement).  I’m not aware of any clean, general way to apply a rectangular transformation to a polar graph or a rotational transformation to a Cartesian graph.  But what I can do is rewrite the polar equations into a parametric form and translate from there.

For 0\le\theta\le 3\pi , r=cos\left(\frac{\theta}{3}\right) becomes \begin{array}{lcl} x_1 &= &cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_1 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array} .  Sliding this \frac{1}{2} a unit to the right makes the parametric equations \begin{array}{lcl} x_2 &= &\frac{1}{2}+cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_2 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array} .

This should align with the standard limaçon, r=\frac{1}{2}+cos\left(\theta\right) , whose parametric equations for 0\le\theta\le 2\pi  are \begin{array}{lcl} x_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot cos\left (\theta\right) \\ y_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot sin\left (\theta\right) \end{array} .

The only problem that remains for comparing (x_2,y_2) and (x_3,y_3) is that their domains are different, but a parameter shift can handle that.

If 0\le\beta\le 3\pi , then (x_2,y_2) becomes \begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\ y_4 &= &cos\left(\frac{\beta}{3}\right)\cdot sin\left (\beta\right) \end{array} and (x_3,y_3) becomes \begin{array}{lcl} x_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left (\frac{2\beta}{3}\right) \\ y_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) \end{array} .

Now that the translation has been applied and both functions operate over the same domain, the two functions must be identical iff x_4 = x_5 and y_4 = y_5 .  It’s time to prove those trig identities!

Before blindly manipulating the equations, I take some time to develop some strategy.  I notice that the (x_5, y_5) equations contain only one type of angle–double angles of the form 2\cdot\frac{\beta}{3} –while the (x_4, y_4) equations contain angles of two different types, \beta and \frac{\beta}{3} .  It is generally easier to work with a single type of angle, so my strategy is going to be to turn everything into trig functions of double angles of the form 2\cdot\frac{\beta}{3} .

\displaystyle \begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\  &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\frac{\beta}{3}+\frac{2\beta}{3} \right) \\  &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot\left( cos\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)-sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\  &= &\frac{1}{2}+\left[cos^2\left(\frac{\beta}{3}\right)\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right) \\  &= &\frac{1}{2}+\left[\frac{1+cos\left(2\frac{\beta}{3}\right)}{2}\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot sin^2\left(\frac{2\beta}{3}\right) \\  &= &\frac{1}{2}+\frac{1}{2}cos\left(\frac{2\beta}{3}\right)+\frac{1}{2} cos^2\left(\frac{2\beta}{3}\right)-\frac{1}{2} \left( 1-cos^2\left(\frac{2\beta}{3}\right)\right) \\  &= & \frac{1}{2}cos\left(\frac{2\beta}{3}\right) + cos^2\left(\frac{2\beta}{3}\right) \\  &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left(\frac{2\beta}{3}\right) = x_5  \end{array}

Proving that the x expressions are equivalent.  Now for the ys

\displaystyle \begin{array}{lcl} y_4 &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\beta\right) \\  &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\frac{\beta}{3}+\frac{2\beta}{3} \right) \\  &= & cos\left(\frac{\beta}{3}\right)\cdot\left( sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+cos\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\  &= & \frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[cos^2 \left(\frac{\beta}{3}\right)\right] sin\left(\frac{2\beta}{3}\right) \\  &= & \frac{1}{2}sin\left(2\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[\frac{1+cos \left(2\frac{\beta}{3}\right)}{2}\right] sin\left(\frac{2\beta}{3}\right) \\  &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) = y_5  \end{array}

Therefore the graph of r=cos\left(\frac{\theta}{3}\right) is exactly the graph of r=\frac{1}{2}+cos\left(\theta\right) slid \frac{1}{2} unit left.  Nice.

If there are any students reading this, know that it took a few iterations to come up with the versions of the identities proved above.  Remember that published mathematics is almost always cleaner and more concise than the effort it took to create it.  One of the early steps I took used the substitution \gamma =\frac{\beta}{3} to clean up the appearance of the algebra.  In the final proof, I decided that the 2 extra lines of proof to substitute in and then back out were not needed.  I also meandered down a couple unnecessarily long paths that I was able to trim in the proof I presented above.

Despite these changes, my proof still feels cumbersome and inelegant to me.  From one perspective–Who cares?  I proved what I set out to prove.  On the other hand, I’d love to know if someone has a more elegant way to establish this connection.  There is always room to learn more.  Commentary welcome.

In the end, it’s nice to know these two polar curves are identical.  It pays to keep one’s eyes eternally open for unexpected connections!