What is the vertex form of #y=4x-x^2#?

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Answer 1 · 2016-01-03T13:54:41

To find the vertex of a parabola we must calculate #{-b}/{2a}# or, equivalently, derive it to find its extreme.

A parabola only has one vertex, which may be a maximum or a minimum. To find it, we must derive it and equal to zero:

#{"d" y}/{"d" x} = 4 - 2 x = 0 rightarrow x = 2#

This is the same that calculating #{-b}/{2a}#, taking that #y=ax^2+bx+c#:

#{-b}/{2a} = {-4}/{2 cdot (-1)} = 2#

Eventually, let us calculate the value of #y# at the vertex:

#y=4 x - x^2 = 4 cdot 2 - 2^2 = 4#

So the vertex of the parabola is at #(2,4)#.