How do you graph $y = 4 csc (x + \frac{π}{2})$ $y=4csc(x+pi/2)$ ?

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Answer 1 · 2018-03-18T11:10:00

As detailed below.

Standard form $y = A sin (B x + C) + D$ $y = A sin(Bx +C) + D$

Given equation is $y = 4 csc (x + (\frac{π}{2}))$ $y = 4 csc (x + (pi/2))$

$Amplitude = None$ $"Amplitude " = "None"$ Function csc x doesn't have amplitude.

$Period = P = \frac{2 π}{| B |} = \frac{2 π}{1} = 2 π$ $"Period " = P = (2pi) / |B| =(2 pi) / 1 = 2pi$

$Phase Shift = - \frac{c}{b} = - \frac{π}{2}$ $"Phase Shift " = -c / b = -pi/2$ #pi/2 " (to the left)"

$Vertical Shift = 0$ $"Vertical Shift" = 0$

graph{4 csc(x + (pi/2)) [-10, 10, -5, 5]}

How do you graph y=4csc(x+π2)y=4csc(x+pi/2)?