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A non-linear system consists of two functions:    f(x)=x^2-2x+2 and g(x)=3x-x^2

Complete the following three parts. Five points each, 15 points total.

 

1. Make a table of values for the functions. The table can be similar to the one below, or it can be vertical like those in the lessons but it must have a minimum of five x-values and the corresponding function values. Include more if you want. Using your table indicate the solution to the system by marking the function values that are equal.

x

 

 

 

 

 

 

 

 

 

f(x)

 

 

 

 

 

 

 

 

 

g(x)

 

 

 

 

 

 

 

 

 

 

2.

Plot a graph of the functions over an interval sufficient to show the solution set. You may carefully sketch or plot your graph manually or use Desmos or other technology. Clearly, indicate and label on the graph the x and y values of the solution(s).

 

 

3.

Solve the system algebraically. (Hint: set the two functions equal to each other.) You should obtain a quadratic equation. Solve it either by factoring or using the quadratic formula. Give the x-values of the solution set, then evaluate the function to find the corresponding y-values. Give the results as ordered pairs of exact values. 

 

Thanks so much for all of your help and support.

Guest Mar 25, 2018
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2+0 Answers

 #1
avatar+6943 
+3

1.  Choose values for  x  and then find  f(x)  and  g(x)  using that  x  value.

 

For the first  x  value I choose  0 . Then...

f(x)  =  f(0)   =   02 - 2(0) + 2  =  2

g(x)  =  g(0)  =  3(0) - 02  =  0

 

x 0        
f(x) 2        
g(x)____ _0_ ___ ___ ___ ___

 

For the second  x  value I choose  1 . Then...

f(x)  =  f(1)  =  12 - 2(1) + 2  =  1 - 2 + 2  =  1

g(x)  =  g(1)  =  3(1) - 12  =  3 - 1  =  2

 

x 0 1      
f(x) 2 1      
g(x)____ _0_ _2_ ___ ___ ___

 

For the third  x  value I choose  1/2 . Then...

f(x)  =  f(1/2)  =  (1/2)2 - 2(1/2) + 2  =  1/4 - 1 + 2  =  1/4 + 1  =  5/4

g(x)  =  g(1/2)  =  3(1/2) - (1/2)2  =  3/2 - 1/4  =  6/4 - 1/4  =  5/4

 

x 0 1 1/2    
f(x) 2 1 5/4    
g(x)____ _0_ _2_ _5/4_ ___ ___

 

For the fourth  x  value I choose  2 . Then...

f(x)  =  f(2)  =  22 - 2(2) + 2  =  4 - 4 + 2  =  2

g(x)  =  g(2)  =  3(2) - 22  =  6 - 4  =  2

 

x 0 1 1/2 2  
f(x) 2 1 5/4 2  
g(x)____ _0_ _2_ _5/4_ _2_ ___

 

Do you see how to do this? I'll let you pick the fifth  x  value.

 

Notice that when  x = 1/2 ,  f(x) = 5/4  and  g(x) = 5/4

So when  x = 1/2 ,  f(x) = g(x)

 

And when  x = 2 ,  f(x) = 2  and  g(x) = 2

So when  x  = 2 ,  f(x) = g(x)

 

The  x  values that make  f(x) = g(x)  are  1/2  and  2 .

So the  x  values that are solutions to the system are  1/2  and  2 .

hectictar  Mar 26, 2018
 #2
avatar+6943 
+3

2.

 

Here's a graph:   https://www.desmos.com/calculator/gpah0jubfb

 

 

3.

 

f(x)  =  x2 - 2x  + 2

g(x)  =  3x - x2

 

Set the two functions equal to each other...

 

x2 - 2x + 2   =   3x - x2

                                         Add  x2  to both sides of the equation.

2x2 - 2x + 2   =   3x

                                         Subtract  3x  from both sides of the equation.

2x2 - 5x + 2   =   0

                                         Split  -5x  into two numbers that multiply to  4  (because 2 * 2 = 4).

2x2 - 4x - 1x + 2  =  0

                                         Factor  2x  out of the first two terms;  Factor  -1  out of the last two terms.

2x(x - 2) - 1(x - 2)  =  0

                                         Factor  (x - 2)  out of both terms.

(x - 2)(2x - 1)  =  0

                                         Set each factor equal to zero.

 

x - 2  =  0          or          2x - 1  =  0

x  =  2               or          2x  =  1

                                        x  =  1/2

 

When  x  =  2  ,  y  =  f(2)  =  g(2)  =  3(2) - 22  =  6 - 4  =  2

When  x = 1/2 ,  y  =  f(2)  =  g(2)  =  3(1/2) - (1/2)2  =  3/2 - 1/4  =  5/4

 

So the solution set is  { (2, 2), (1/2, 5/4) }

hectictar  Mar 26, 2018

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