What is the vertex form of $y = 4 x - x^{2}$ $y=4x-x^2$ ?

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Answer 1 · 2016-01-03T13:54:41

To find the vertex of a parabola we must calculate $\frac{- b}{2 a}$ ${-b}/{2a}$ 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:

$\frac{d y}{d x} = 4 - 2 x = 0 \to x = 2$ ${"d" y}/{"d" x} = 4 - 2 x = 0 rightarrow x = 2$

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

$\frac{- b}{2 a} = \frac{- 4}{2 \cdot (- 1)} = 2$ ${-b}/{2a} = {-4}/{2 cdot (-1)} = 2$

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

$y = 4 x - x^{2} = 4 \cdot 2 - 2^{2} = 4$ $y=4 x - x^2 = 4 cdot 2 - 2^2 = 4$

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

What is the vertex form of y=4x-x^2y=4x−x2y=4x-x^2?