Can someone explain how to answer this sine graph question please?

Question

The function is given by f(x) = 2 + 5 sin 3x for 0^o <= x <= 180^o

1. State the amplitude and period of f(x)

2.Draw the graph of y = f(x)

I already know the amplitude is 5 but how do i find the period and draw that graph :S please help ty

Answer 1

Here is the graph for  0 ≤ x ≤ π :

http://tinyurl.com/69xtph

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Answer 2

The period of y = sin (bx) is:

(2π) / | b |

y = 2 + 5sin (3x)

Period = (2π) / | 3 |

Period = (2/3)π

Amplitude = 5

The two is a horizontal shift, meaning that the entire graph of the function will shift up by two. So make the y-axis (-7, 7).

The period is (2/3)π. This means that one oscillation will take place within this period. If I&#039;m reading your question correctly, you are asked to find half of the period (π = 180, 2π = 360).

Label the x-axis from 0 to (2/3)π.

When x is 0, y = 2.

When x is π/6, y = 7

When x is (1/3)π, y = 2

When x is (π/6), y = -3

When x is (2/3)π, y = 2

Plot the points and draw the function.

I included a link to the graph below. The sin function in red in the function with the horizontal shift. The function in blue is without the horizontal shift. I just put it there to compare.

Answer 3

1. Let T be the period, then

f(x)=f(x+T)=&gt;

sin3(x+T)=sin3x=&gt;

sin(3T/2)=0=&gt;

T=2pi/3 (pi=3.141592...)

2) For drawing the graph of f(x), you should displace the graph

of y=5sin3x up by 2 units from the x-axis.