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

Feb 13, 2018

The vertex is at the point $\left(2 , 4\right)$.

#### Explanation:

If you write your quadratic in standard form, ($a {x}^{2} + b x + c$) then the $x$-coordinate of the vertex is calculated by $\frac{- b}{2 a}$.

Let's convert that equation to standard form:

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

$\quad = - {x}^{2} + 4 x$

$\quad = - 1 {x}^{2} + 4 x + 0$

In this case, our $a$ value is $- 1$, the $b$ value is 4, and the $c$ value is $0$. This means the $x$-coordinate of the vertex is $\frac{- 4}{2 \left(- 1\right)}$ which is $2$.

Now, to find the $y$-coordinate, simply plug $2$ into the equation and see the value that it returns:

$y = - 1 {x}^{2} + 4 x + 0$

$\quad \implies - 1 {\left(2\right)}^{2} + 4 \left(2\right) + 0$

$\quad = - 1 \cdot 4 + 8$

$\quad = - 4 + 8$

$\quad = 4$

Now we know that the vertex is at an $x$ of $2$ and a $y$ of $4$, or the point $\left(2 , 4\right)$.

You can check this by graphing the parabola: graph{4x-x^2 [-8, 12, -2, 8]}