CPhill

6. The period is  pi / 2

7.  The graph is vertically stretched by a factor of 2  and shifted up 3 units

5.   

We can solve for the period, p,  as follows

(1/4) p  = 2pi      multiply both sides by 4

p   =  8 pi

The frequency  =  1 / period  =    1  / [ 8 pi ]  

4.      f(x)  =  (1/2)cos ( x)  +  5

The "1/2 "  does not affect the midline..it only halves the normal amplitude

The "5" shifts  the normal cosine graph up 5 uints

So....the midline  is   y = 5

Here's the graph : https://www.desmos.com/calculator/lkmkkpvrzq

3.

Reflected over the x axis gives us  y  = -cos (x)

Shifted down 2 units gives  y  =  -cos(x) - 2 = g(x)

1.  Minimum value  is   2.5

2.  Period  =  6pi      Amplitude = 5      Midline :  y  = -2 

Since we have a y intercept of  (0, -2)  the sine function seems he best choice...here's the function :

y  =  5sin (x /3 )  - 2

Here's the graph : https://www.desmos.com/calculator/shtgla34yy

OK.....thanks for your explanation, Melody  !!!

OK....!!!

Good to see you, RedGlasses !!!

Very nice, hectictar...!!!

1.  The first graph is just the sine function shifted down two units

Here's  the graph of both :   https://www.desmos.com/calculator/pl7ix8zvkb

2.. This is just the cosine graph  with a period 1/2 as long as  the parent function and shifted up by 1 unit

Here's the graph of both: https://www.desmos.com/calculator/nokkptgkli

3.  A little difficult, but here's one possibility...

In  the first second  the pendulum  swings to the max position (to the right) 

In the next second, it returns to its neutral position

In the third second, it swings to min position (to the left )

In the last second, it returns to the neutral position again

So....the function is

y  =  6sin (pi * t / 2)

Here's the graph : https://www.desmos.com/calculator/0cslmqnaad