How do you graph $y = 1 - cos (π x)$ $y = 1 - cos(pix)$ ?

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Answer 1 · 2018-08-07T14:21:02

As below.

$y = 1 - cos (π x)$ $y = 1 - cos (pi x)$

$y = A cos (B x - C) + D$ $y = A cos (Bx - C) + D$ is the standard form of cosine function.

$A = - 1, B = π, C = 0, D = 1$ $A = -1, B = pi, C = 0, D = 1$

$A m p l i t u d e = | A | = | - 1 | = 1$ $Amplitude = |A| = |-1| = 1$

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

Phase Shift $= - \frac{C}{B} = 0$ $= -C / B = 0$

Vertical Shift $= D = 1$ $= D = 1$

graph{- cos(pi x) + 1 [-10, 10, -5, 5]}

How do you graph y=1−cos(πx)y = 1 - cos(pix)?