+0  
 
+1
45
6
avatar+1672 

4.) Sketch a graph of a polynomial that would have:


a.) 7 real zeros
b.) 5 real and 2 complex zeros
c.) 4 complex zeros
d.) 2 complex and 4 real zeros

 Feb 10, 2019
 #1
avatar+98060 
+2

a)  We can make up whatever we want

For the first....a possibility is    ( x - 3)^5 (x - 2)^2

[ We have repeated real zeroes, but the problem doesn't say that every zero has to be unique ]

The graph of this is here :   https://www.desmos.com/calculator/td0qq1sakf

 

b)  We can have

(x  - 3) ( x + 3) ( x - 4) ( x + 4) (x + 6) ( x^2 + 1)

The graph of this is here : https://www.desmos.com/calculator/0f8x7e1irp

 

c) We can have

(x^2 + 1) ( x^2 + 3)

Graph : https://www.desmos.com/calculator/sqj7sk0rcf

 

d) We can have

(x^2 + 1) (x + 3) ( x - 3) ( x + 2) ( x - 2)

Graph : https://www.desmos.com/calculator/nyabj6mmtz

 

I hope these links work, GM !!!

 

cool cool cool

 Feb 10, 2019
 #2
avatar+1672 
+1

Thanks so much, CPhill! I wouldn't even know how to make polynomial with 7 real zeros.

Also, can you help me solve this problem? https://web2.0calc.com/questions/working-together-formula-3

I got my work written down but I want to make sure of if it is correct.

 

-- 7H3_5H4D0W

 Feb 10, 2019
edited by GAMEMASTERX40  Feb 10, 2019
 #3
avatar+98060 
+2

OK....using your formula from last week...we have

 

1/3 + 1/4  =  1/ Total Time Working Together

 

4/12  + 3/12 =  1 / T

 

7 /12  = 1 / T       [ remember that we can write ]

 

12 / 7  =  T / 1

 

12/ 7  = T     

 

T = 12/7 hrs ≈     1.7 hrs  =    1 + .7(60)  =   1 hr 42 min

 

 

 

cool cool cool

CPhill  Feb 10, 2019
 #4
avatar+1672 
+1

Nice, my work makes sense now. Thanks!

 

-- 7H3_5H4D0W

GAMEMASTERX40  Feb 10, 2019
 #5
avatar+98060 
+1

OK.....Mr. "SHADOW"

 

 

cool cool cool

CPhill  Feb 10, 2019
 #6
avatar+1672 
0

Mr. SHADOW... ok then.

GAMEMASTERX40  Feb 10, 2019

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