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How do you find the amplitude and period of #y=2sec(1/4t)#?

Answer 1

Aaliyah Abdullah

Amplitude = None, Period #= 8pi#

Standard form #y = a sec(bx - c) + d#

Given equation is #y = 2 sec(1/4)t#

#"Amplitude = None"#

#"Period" = (2pi)/|b| = (2pi) / (1/4) = 8pi#

graph{2 sec((1/4)x) [-10, 10, -5, 5]}

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Answer 2

Carter Anderson

To find the amplitude and period of (y = 2\sec\left(\frac{1}{4}t\right)), note that the amplitude of a secant function is the absolute value of the coefficient of the secant term, and the period is (2\pi) divided by the absolute value of the coefficient of (t) inside the function.

For the given function (y = 2\sec\left(\frac{1}{4}t\right)):

The amplitude is (2).
The period is (2\pi \times 4 = 8\pi).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.