Express 6+4x-x^2 in the form a-(x+b)^2, where a and b are intergers. (i) find the coordinates of the turning point of the curve y=6+4x - x^2 and determine the nature of this turning point. the function f is defined by f:x - 6+4x-x^2 for the domain 0

Question

Express 6+4x-x^2 in the form a-(x+b)^2, where a and b are intergers. (i) find the coordinates of the turning point of the curve y=6+4x - x^2 and determine the nature of this turning point. the function f is defined by f:x -> 6+4x-x^2 for the domain 0

anonymous · Answer

6+4x-x^2=-(x^2-4x-6)=-{(x-2)^2 -10}=10-(x-2)^2

anonymous · Answer

The turning point is also called stationary point. You can find it by solving f'(x)=0
where f'(x) is the derivative of the function

anonymous · Answer

so f'(x)=4-2x
4-2x=0
x=2

anonymous · Answer

This is a local maximum as x^2 has a - sign so it is a parabola that is facing down

anonymous · Answer

I.. don't understand ;A; -feels stupid-

anonymous · Answer

The range can be found by f(2)=10 
That is the maximum so the function takes all values from -infinity until 10
$$(-\infty;10$$

anonymous · Answer

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anonymous · Answer

Sorry I do not have time now to explain as I have to go to work :(

anonymous · Answer

it's okay... thanks :)

anonymous · Answer

you get this?

anonymous · Answer

not at all. my brain is being incredibly slow today. like any other days.. >_> i don't know how a vertex matter............