Question

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.

Answer 1

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 :   $\frac{2 \pi}{|b|} = 5$

            ⇒ $b = \frac{2\pi}{5}$

Phase shift :  $\frac{c}{b} = 2.5$

                   $\Rightarrow \frac{c}{2 \pi/5}=\frac{5}{2}$

                  ∴ $c = - \pi$

The vertical displacement : d = 3 units up

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

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

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

The graph is attached below.