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

Jan 3, 2016

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

#### Explanation:

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:

$\left\{\text{d" y}/{"d} x\right\} = 4 - 2 x = 0 \rightarrow x = 2$

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

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

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 $\left(2 , 4\right)$.