How do you find the amplitude and period of $y = 2 sec (\frac{4}{5} θ)$ $y=2sec(4/5theta)$ ?

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How do you find the amplitude and period of $y = 2 sec (\frac{4}{5} θ)$ $y=2sec(4/5theta)$ ?

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Answer 1 · 2017-06-24T01:22:19

Amplitude is $2$ $2$ .
Period is $\frac{5 π}{2}$ $(5pi)/2$ or $7.85398163$ $7.85398163$ .

Explanation:

Standard Form: $y = a sec b (x - c) + d$ $y=asecb(x-c)+d$
Equation: $y = 2 sec (\frac{4}{5} θ)$ $y=2sec(4/5theta)$
$a = 2$ $a=2$
$b = \frac{4}{5}$ $b=4/5$
$c =$ $c=$ none
$d =$ $d=$ none

AMPLITUDE
amplitude = distance from middle of graph (y=0) to highest and lowest points on graph
formula: $| a |$ $|a|$
$| 2 |$ $|2|$
$2$ $2$
Amplitude is $2$ $2$ .

PERIOD
period = interval in which the graph repeats itself
formula: $\frac{2 π}{| b |}$ $(2pi)/|b|$
$\frac{2 π}{∣ ∣ \frac{4}{5} ∣ ∣}$ $(2pi)/|4/5|$
$\frac{2 π}{\frac{4}{5}}$ $(2pi)/(4/5)$
$2 π \div \frac{4}{5}$ $2pi-:4/5$
$2 π \cdot \frac{5}{4}$ $2pi*5/4$
$\frac{10 π}{4}$ $(10pi)/4$
$\frac{5 π}{2} = 7.85398163$ $(5pi)/2 = 7.85398163$
Period is $\frac{5 π}{2}$ $(5pi)/2$ or $7.85398163$ $7.85398163$ .

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