# How do you find the amplitude, period, phase shift given y=2csc(2x-1)?

Apr 18, 2018

The $2 x$ makes the period $\pi$, the $- 1$ compared to $2$ in $2 x$ makes the phase shift $\frac{1}{2}$ radian, and the divergent nature of cosecant makes the amplitude infinite.

#### Explanation:

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Graph of $2 \csc \left(2 x - 1\right)$

graph{2 csc(2x - 1) [-10, 10, -5, 5]}
The trig functions like $\csc x$ all have period $2 \setminus \pi .$ By doubling the coefficient on $x$, that halves the period, so the function $\csc \left(2 x\right)$ must have a period of $\pi$, as must $2 \csc \left(2 x - 1\right)$.

The phase shift for $\csc \left(a x - b\right)$ is given by $\frac{b}{a} .$ Here we have a phase shift of $\frac{1}{2}$ radian, approximately ${28.6}^{\setminus} \circ$. The minus sign means $2 \csc \left(2 x - 1\right)$ leads $2 \csc \left(2 x\right)$ so we call this a positive phase shift of $\frac{1}{2}$ radian.

$\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$ so it diverges twice per period. The amplitude is infinite.