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0
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ty to the guy who answered!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Guest May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
 #1
avatar+91027 
+1

Here's a few

 

Task 2

 

y  = x^2       (quadratic)

y = x            (linear)

 

Sub  the   second equation into the first for y

 

x  = x^2       subtract x from  both sides

 

0  = x^2  - x      

x^2  - x   =  0   factor

x ( x - 1)  =0

Set each factor to 0  and solve for  x

 

x  = 0        x  - 1  =  0   

                add 1 to  both sides

 

x = 0          x  = 1

 

So....when x  = 0, y  = 0

And when x  = 1, y  = 1

So....the solutions are  (0,0)  and (1,1)

 

Here's a graph ;

 

https://www.desmos.com/calculator/ldpimov7kd

 

cool cool cool

CPhill  May 2, 2018
 #2
avatar
0

ty very much

Guest May 2, 2018
 #3
avatar+91027 
0

Task 3

 

y  =   (x + 3)

        ______

         (x - 2) 

 

The horizontal asymptote  occurs at  y  = 1

The vertcal asymptote occurs at  x  = 2

The x intercept is  at  x  = -3

The y intercept is  at   (-3/2)

 

Here's a  graph : 

 

https://www.desmos.com/calculator/4l3egciy58

 

cool cool cool

CPhill  May 2, 2018
 #4
avatar
+1

thanks again

Guest May 2, 2018
 #5
avatar+91027 
+1

Task 5

 

Let  the polynomial  be    x^2  - 4x  + 4

Let the  linear binomial be    ( x - 2)

 

 

Part 1

 

               x  -  2

x - 2   [   x^2  -  4x  + 4  ]

              x^2   -2x

             _____________

                       -2x   + 4

                       -2x   + 4

                       _______

                                 0

 

Part 2

 

f(a)   = f(2)  =   2^2  - 4(2)  + 4   =    4 - 8  + 4   =  0

 

Part 3 

 

Since  we obtained a "0"  with polynomial long division......then x  = 2   is  a  root. The Remainder Theorem says that if f(a)  = 0, then  "a" is a root.  This is borne out by the fact that f(2) = 0  where  "a"   = 2.

 

 

 

cool cool cool

CPhill  May 2, 2018
 #6
avatar+91027 
+1

Task 6

 

Let  N  = 37

We can write this as  (x + y)  =  (49-  12)

 

So....note that

 

(49 -  12) ^2  =    49^2  - 2(49*12) + 12^2  =   2401  - 1176  + 144    = 1369  =  372 

 

 

cool cool cool

CPhill  May 2, 2018
 #7
avatar
+1

helpful, thank you, you helped me learn a bit too :D

Guest May 2, 2018
 #8
avatar+91027 
+1

Task 4

 

x + a      =    b

____           ___

  ax               x

 

We can "engineer"  an extraneous solution, as follows :

 

Cross-multiply ;

 

x (x + a)  =  bax    simplify

 

x^2  + ax  = bax

 

x^2  + ax - bax  = 0              let  a  =  1, let  b  = 5....so we have

 

x^2  + 1x  - (5)(1)(x)  = 0

 

x^2 + 1x  - 5x  = 0

 

x^2  - 4x  = 0      factor

 

x(x  - 4)  = 0

 

Set both factors to 0   and solve for  x

 

x  = 0           x  - 4  =0

                      x  = 4

 

So.....note that the original problem becomes ;

 

x + 1            5

____   =      __

   1x              x

 

Note that if  x  = 4

 

4 + 1             5

_____   =    ___       which is true

   1*4             4     

 

 

 

However...if x  =  0   ....both denominators are undefined because we cannot divide  by 0....thus....x  = 0  is an extraneous solution....!!!!!

 

 

 

cool cool cool

CPhill  May 2, 2018

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