# How do you find the vertex of this parabola y=4x-x^2?

##### 2 Answers
Apr 17, 2015

Vertex (2,4)

Rewrite the equation as y = - $\left({x}^{2} - 4 x\right)$
= -$\left({x}^{2} - 4 x + 4\right)$ +4
= - ${\left(x - 2\right)}^{2}$+4

The vertex is (2,4)

Apr 17, 2015

Vertex is $\left(2 , 4\right)$

Step 1: Complete the square

$y = - \left[{x}^{2} - 4 x\right] = - \left[{\left(x - 2\right)}^{2} - 4\right]$

Step 2: Arrange so that you get the form ${\left(x - {x}_{v}\right)}^{2} = 4 a \left(y - {y}_{v}\right)$

$y = - \left[{\left(x - 2\right)}^{2} - 4\right] = - {\left(x - 2\right)}^{2} + 4$

$\implies {\left(x - 2\right)}^{2} = 4 - y$

$\implies {\left(x - 2\right)}^{2} = - \left(y - 4\right)$

From here you can conclude that the vertex is at $\left(2 , 4\right)$