+0  
 
0
90
1
avatar+166 

So, another question. 

 

But when I have a system of inequalities, which points do I count as part of the system and which points do I not. For example, if I had y>x2 and y<-2(x+2)2 +6 which points would be considered solutions

MrPatel  Apr 30, 2018
 #1
avatar+87333 
+1

y > x^2

y < -2(x + 2)^2 + 6

 

Instead of inequalities, let's just set these equal and see if we can find the correct solutions

 

x^2  = -2(x + 2)^2  + 6

 

x^2 = -2(x^2 + 4x + 4) + 6

 

x^2 = -2x^2 -8x - 8  + 6   

 

x^2 = - 2x^2 - 8x - 2      rearrange as

 

3x^2 + 8x + 2  =   0

 

Solving this with the Quadratic Formula,  we get the solutions :

 

x = ( -4 - √10 )  / 3  ≈   -2.387

 

x = (-4 + √10 ) / 3  ≈ -0.279

 

We have  three possible intervals  for solutions

 

(-infinity, ≈ -2.87 )    U ( ≈-2.87, ≈ -0.279) U  (≈ -0.279, infinity)

 

The way these usually work is that either the two outside intervals work or the middle interval does

 

Here's a graph to prove that the middle interval is correct  :

 

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

 

The solution area is where both graphs overlap

 

 

 

coolcool cool

CPhill  May 1, 2018

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