How do you graph $y = - 3 cot (\frac{1}{2} (x + π)) - 1$ $y=-3cot(1/2(x+pi))-1$ ?

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How do you graph $y = - 3 cot (\frac{1}{2} (x + π)) - 1$ $y=-3cot(1/2(x+pi))-1$ ?

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Answer 1 · 2018-06-16T10:01:12

As below.

Explanation:

$Standard form of cotangent function y = A cot (B x - C) + D$ $"Standard form of cotangent function " y = A cot (Bx -C) + D$

$y = - 3 cot g r a p h {- 3 cot (\frac{x}{2} + \frac{π}{2}) - 1 [- 10, 10, - 5, 5]} (\frac{x}{2} + \frac{π}{2}) - 1$ $y = -3cot graph{-3 cot(x/2 + pi/2) - 1 [-10, 10, -5, 5]} (x/2 + pi/2) -1$

$A = - 3, B = \frac{1}{2}, C = - \frac{π}{2}, D = - 1$ $A = -3, B = 1/2, C = -pi/2, D = -1$

$A m p l i t u d e = | A | = NONE for cotangent function$ $Amplitude = |A| = "NONE for cotangent function"$

$P e r i o d = \frac{π}{| B |} = \frac{π}{\frac{1}{2}} = 2 π$ $"Period = pi / |B| = pi / (1/2) = 2pi$

$Phase Shift = - \frac{C}{B} = \frac{\frac{π}{2}}{\frac{1}{2}} = π, π to the right$ $"Phase Shift " = -C / B = (pi/2)/(1/2) = pi, pi " to the right"$

$Vertical Shift = D = - 1$ $"Vertical Shift " = D = -1$

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How do you graph $y = - 3 cot (\frac{1}{2} (x + π)) - 1$ $y=-3cot(1/2(x+pi))-1$ ?

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Explanation:

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