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 = x = 5. (ii) find the range of f
6+4x-x^2=-(x^2-4x-6)=-{(x-2)^2 -10}=10-(x-2)^2
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
so f'(x)=4-2x 4-2x=0 x=2
This is a local maximum as x^2 has a - sign so it is a parabola that is facing down
I.. don't understand ;A; -feels stupid-
The range can be found by f(2)=10 That is the maximum so the function takes all values from -infinity until 10 \[(-\infty;10\]
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Sorry I do not have time now to explain as I have to go to work :(
it's okay... thanks :)
you get this?
not at all. my brain is being incredibly slow today. like any other days.. >_> i don't know how a vertex matter............
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