Question

question 24
$y = -2\tan(4x - \pi) - 1$, what is the period of this function?
(a) 2 \quad (b) $-\frac{\pi}{4}$ \quad (c) $\frac{\pi}{2}$ \quad (d) -2 \quad (e) $\frac{\pi}{4}$ \quad (f) $-\frac{\pi}{2}$
\bigcirc a
\bigcirc b
\bigcirc c
\bigcirc d
\bigcirc e
\bigcirc f

Explanation:

First Question (Phase Shift of \( y = -2\cos(4x + \pi) - 1 \))

Step1: Recall the general form of cosine function

The general form of a cosine function is \( y = A\cos(Bx - C) + D \), where the phase shift is \( \frac{C}{B} \).

Step2: Rewrite the given function in general form

Given \( y = -2\cos(4x + \pi) - 1 \), we can rewrite it as \( y = -2\cos(4x - (-\pi)) - 1 \). Here, \( B = 4 \) and \( C = -\pi \).

Step3: Calculate the phase shift

Using the formula for phase shift \( \frac{C}{B} \), we substitute \( C = -\pi \) and \( B = 4 \). So, the phase shift is \( \frac{-\pi}{4}=-\frac{\pi}{4} \).

Step1: Recall the period formula for tangent function

The general form of a tangent function is \( y = A\tan(Bx - C) + D \), and the period of \( \tan(x) \) is \( \pi \), so the period of \( y = A\tan(Bx - C) + D \) is \( \frac{\pi}{|B|} \).

Step2: Identify \( B \) from the given function

For the function \( y = -2\tan(4x - \pi) - 1 \), we have \( B = 4 \).

Step3: Calculate the period

Using the period formula \( \frac{\pi}{|B|} \), substitute \( B = 4 \). So, the period is \( \frac{\pi}{4} \)? Wait, no, wait. Wait, the standard period of \( \tan(kx) \) is \( \frac{\pi}{|k|} \). Wait, in the function \( y=-2\tan(4x - \pi)-1 \), the coefficient of \( x \) is \( 4 \), so \( B = 4 \). Then the period is \( \frac{\pi}{|4|}=\frac{\pi}{4} \)? Wait, no, wait, no. Wait, the tangent function \( \tan(x) \) has a period of \( \pi \). When we have \( \tan(Bx) \), the period is \( \frac{\pi}{|B|} \). So for \( B = 4 \), the period is \( \frac{\pi}{4} \)? Wait, but let's check the options. Wait, the options are (a) 2, (b) \( -\frac{\pi}{4} \), (c) \( \frac{\pi}{2} \), (d) -2, (e) \( \frac{\pi}{4} \), (f) \( -\frac{\pi}{2} \). Wait, maybe I made a mistake. Wait, no, the standard period of \( \tan(x) \) is \( \pi \), so for \( \tan(4x) \), the period is \( \frac{\pi}{4} \)? Wait, no, wait, no. Wait, the formula for the period of \( \tan(Bx + C) \) is \( \frac{\pi}{|B|} \). So if \( B = 4 \), then period is \( \frac{\pi}{4} \), which is option (e). Wait, but let's re - check. Let's take \( y = \tan(x) \), period \( \pi \). For \( y=\tan(2x) \), the period is \( \frac{\pi}{2} \), because \( \tan(2(x+\frac{\pi}{2}))=\tan(2x + \pi)=\tan(2x) \). So in general, for \( y = \tan(Bx) \), period is \( \frac{\pi}{|B|} \). So for \( B = 4 \), period is \( \frac{\pi}{4} \). So the answer is (e).

Step1: Recall the period formula for tangent function

The period of the tangent function \( y = A\tan(Bx - C)+D \) is given by \( \frac{\pi}{|B|} \), where \( B \) is the coefficient of \( x \) inside the tangent function.

Step2: Identify the value of \( B \)

For the function \( y=-2\tan(4x - \pi)-1 \), the coefficient of \( x \) (i.e., \( B \)) is \( 4 \).

Step3: Calculate the period

Using the formula \( \frac{\pi}{|B|} \), substitute \( B = 4 \). We get \( \frac{\pi}{|4|}=\frac{\pi}{4} \).

Answer:

b. \( -\frac{\pi}{4} \)

Second Question (Period of \( y = -2\tan(4x - \pi) - 1 \))