## By what factor is the graph of the trig function stretched?

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
chiefboo
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### By what factor is the graph of the trig function stretched?

y= 2 - (5/2) sec( (2pi/5)x-4pi)

By what factor is the graph of this secant function being stretched horizontally?

I thought you're supposed to divide 2pi by 2pi/5:

2pi/(2pi/5)=

2pi * 5/2pi =

10pi/2pi = 5

Please help me understand this. I have a unit exam next week and a quiz I need to be studying for tomorrow.

jg.allinsymbols
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### Re: By what factor is the graph of the trig function stretch

Yes it is stretched or DIVIDED by 2*pi/5.

Actually squeezed by factor of 2*pi/5. The period is shorter.

stapel_eliz
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chiefboo wrote:y= 2 - (5/2) sec( (2pi/5)x-4pi)

By what factor is the graph of this secant function being stretched horizontally?

I think the confusion here lies in that you're (quite sensibly) trying to find the period, while the question is asking "by what factor is the [period] being stretched?" So you're correct that the period is "5", but this period is created by "stretching" the regular period of $2\pi$ by a "factor" of $2\pi / 5$.

That is, you did the following:

. . . . .$\mbox{new period: }\, \frac{2\pi}{B}\, =\, \frac{2\pi}{\left(\frac{2\pi}{5}\right)}\, =\, 5$

...where "B" is the value multiplied against the variable inside the trig function. On the other hand, I think the question is looking for:

. . . . .$\mbox{factor: }\, 5\, =\, \frac{(2\pi)(5)}{2\pi}\, =\, \frac{2\pi}{\left(\frac{2\pi}{5}\right)}\, =\, \frac{2\pi}{F}$

...where "F" is the "factor" they're wanting.

Yes, it's confusing.