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How do I solve this y=x-x^2?

 

I need to know the turning points. Idk if it is the right word but i hope you understand me!

 May 26, 2016

Best Answer 

 #1
avatar+118613 
+5

How do I solve this y=x-x^2?

 

I need to know the turning points. Idk if it is the right word but i hope you understand me!

 

This is a parabola, I know that because of the x^2 term

This is a concave down.  I know that because of the - sign in front of the x^2 term.

 

\(y=x-x^2\\ y=-x^2+x\\ y=-x(x-1)\\\)

 

The roots are  x=0  and x=1

the axis of symmetry  is x= 0.5

 

when x=0.5

y= -0.5 * (0.5-1)

y= -0.5 * -0.5

y= 0.25

 

So the tuning point is a maximum and it is at     (0.5, 0.25)

 

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

 May 26, 2016
 #1
avatar+118613 
+5
Best Answer

How do I solve this y=x-x^2?

 

I need to know the turning points. Idk if it is the right word but i hope you understand me!

 

This is a parabola, I know that because of the x^2 term

This is a concave down.  I know that because of the - sign in front of the x^2 term.

 

\(y=x-x^2\\ y=-x^2+x\\ y=-x(x-1)\\\)

 

The roots are  x=0  and x=1

the axis of symmetry  is x= 0.5

 

when x=0.5

y= -0.5 * (0.5-1)

y= -0.5 * -0.5

y= 0.25

 

So the tuning point is a maximum and it is at     (0.5, 0.25)

 

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

Melody May 26, 2016
 #2
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+5

The graph.

 

 May 26, 2016

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