55.6k views
3 votes
Amplitude of 16 units, a period of 5 units, a vertical displacement of 3 units up and a phase shift of 2.5 units left.

User Ayudh
by
8.3k points

1 Answer

2 votes

The question is incomplete. The complete question is :

A sine function that has an amplitude of 16 units, a period of 5 units,a vertical displacement of 3 units up and a phase shift of 2.5 units left. Graph this function and show each step.

Solution :

The standard form of the sine function is

y = a sin (bx - c) + d

Here given :

Amplitude, a = 16

Period :
$(2 \pi)/(|b|) = 5$


$b = (2\pi)/(5)$

Phase shift :
$(c)/(b) = 2.5$


$\Rightarrow (c)/(2 \pi/5)=(5)/(2)$


$c = - \pi$

The vertical displacement : d = 3 units up

Now substituting a, b, c and d values in the standard form gives :


$y = 16 \sin \left( (2 \pi)/(5)x -(- \pi)\right) + 3$


$y = 16 \sin \left( (2 \pi)/(5)x + \pi\right) + 3$

The graph is attached below.

Amplitude of 16 units, a period of 5 units, a vertical displacement of 3 units up-example-1
User Mechnicov
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories