We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
327
5
avatar

1.

2.

3.

4.

5.

6.

7.

 Jan 29, 2018
 #1
avatar+102372 
0

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

 

 

cool cool cool

 Jan 29, 2018
 #2
avatar+102372 
0

3.

 

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

 

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

 

 

cool cool cool

 Jan 29, 2018
 #3
avatar+102372 
0

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

 

 

cool cool cool

 Jan 29, 2018
 #4
avatar+102372 
0

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 ]  

 

 

cool cool cool

 Jan 29, 2018
 #5
avatar+102372 
0

6. The period is  pi / 2

 

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

 

 

cool cool cool

 Jan 29, 2018

7 Online Users

avatar