Solve the problem.
-For the equation $y = - \frac { 1 } { 2 } \sin ( 4 x + 3 \pi$ ), identify (i) the amplitude, (ii) the phase shift, and (iii) the period.
A) (i) $- \frac { 1 } { 2 }$
(ii) $- \frac { 4 \pi } { 3 }$
(iii) 4

B) (i) $\frac { 1 } { 2 }$
(ii) $- \frac { 3 \pi } { 4 }$
(iii) $\frac { \pi } { 2 }$

C) (i) 2
(ii) $3 \pi$
(iii) $\frac { \pi } { 2 }$

D) (i) $\frac { 1 } { 2 }$
(ii) $- \frac { 3 \pi } { 4 }$
(iii) 4

Question

Solve the problem.
-For the equation $y = - \frac { 1 } { 2 } \sin ( 4 x + 3 \pi$ ), identify (i) the amplitude, (ii) the phase shift, and (iii) the period.
A) (i) $- \frac { 1 } { 2 }$
(ii) $- \frac { 4 \pi } { 3 }$
(iii) 4

B) (i) $\frac { 1 } { 2 }$
(ii) $- \frac { 3 \pi } { 4 }$
(iii) $\frac { \pi } { 2 }$

C) (i) 2
(ii) $3 \pi$
(iii) $\frac { \pi } { 2 }$

D) (i) $\frac { 1 } { 2 }$
(ii) $- \frac { 3 \pi } { 4 }$
(iii) 4

Quizplus · Accepted Answer

The amplitude of a sine function is the absolute value of the coefficient in front of the sine, which is $|\frac{1}{2}|$. The phase shift is calculated by solving $4x + 3\pi = 0$ for $x$, giving $x = -\frac{3\pi}{4}$. The period of a sine function is $\frac{2\pi}{|B|}$, where $B$ is the coefficient of $x$ inside the sine function, so here it is $\frac{2\pi}{4} = \frac{\pi}{2}$.

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Solve the Problem $y = - \frac { 1 } { 2 } \sin ( 4 x + 3 \pi$