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March 8th, 2015, 01:38 PM   #1
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Question plotting sec and csc

I'm a relatively new math tutor, and I have a question about the best way to explain how to plot transformed sec and csc functions. Really this could be about sin and cos and tan as well, but I'll stick with sec in this question.

Let's say I have a function of the form

f(x) = a (sec (bx + c)) + d

We know the asymptotes of the untransformed sec function are at -3 pi /2 , -pi/2, pi/2, 3 pi /2, etc. And sec is positive between -pi/2 and pi/2, negative between pi/2 and 3pi/2, etc.

So I could explain this two ways.

1. I could find the period and phase shift. The period is 2pi/b. The phase shift you get by factoring out the b, so you get sec(b(x + c/b)) so the phase shift is -c/b. Then I multiply each of the x values of the asymptotes by 1/b and subtract c/b. I track which intervals have positive y-values and which have negative.

2. I could solve the following equations:

pi/2 = b * x_1 + c
3pi/2 = b * x_2 + c
...

etc. for each of the asymptotes of the untransformed sec function. Then my new asymptotes are at x_1, x_2, etc.

These are obviously the same thing in the end. However, I like (2) because I think it might be simpler to see why it works.

Any advice?

Mike
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March 8th, 2015, 01:51 PM   #2
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Quote:
Let's say I have a function of the form

f(x) = a (sec (bx + c)) + d
I would graph f(x) = a (cos (bx + c)) + d , then plot the secant using what I call the "hump to hump" method ... example attached.

$\displaystyle f(x) = 2\cos[3(x+4)]+5$

$\displaystyle \color{red}{f(x) = 2\sec[3(x+4)]+5}$
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File Type: jpg cos_sec_graf.jpg (19.9 KB, 0 views)
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March 8th, 2015, 02:05 PM   #3
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Thanks, that makes sense. But if I plot cos then I have to find the zero crossings rather than the asymptotes, so my question is pretty much the same. Should I teach it by finding the period, then the phase shift? Or by solving a series of equations to find the asymptotes? Or both methods?
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March 8th, 2015, 02:49 PM   #4
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I like to use the period and shift ... they learn all those transformations for a reason.

Let's say you wanted to graph $y = 2\sec[3(x+4)]+5$ ...

start with a sketch of $\cos(3x)$ ... period is $\frac{2\pi}{3}$ and has zeros at $\frac{\pi}{6}$ and $\frac{3\pi}{6}$ in the interval $0 \le x \le \frac{2\pi}{3}$, and it's rather simple to see all zeros for each additional period will be at odd multiples of $\frac{\pi}{6}$ ...

move on to $\cos[3(x+4)]$ where the zeros are translated 4 units left ...

so, each zero would be at $x = -4 + \frac{(2k+1)\pi}{6} \, ; \, k \in \mathbb{Z}$

which is also where each vertical asymptote for the secant function will be.
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March 8th, 2015, 06:38 PM   #5
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Thanks! Makes sense.
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