# What is the vertex of y= -x^2+40x-16?

Aug 1, 2018

The vertex is at $\left(20 , 384\right)$.

#### Explanation:

Given: $y = - {x}^{2} + 40 x - 16$

This equation is in standard quadratic form $\left(y = a {x}^{2} + b x + c\right)$, meaning we can find the $x$-value of the vertex using the formula $\frac{- b}{2 a}$.

We know that $a = - 1$, $b = 4$, and $c = - 16$, so let's plug them into the formula:
$x = \frac{- 40}{2 \left(- 1\right)} = 20$

Therefore, the $x$-coordinate is $20$.

To find the $y$-coordinate of the vertex, plug in the $x$-coordinate and find $y$:
$y = - {x}^{2} + 40 x - 16$

$y = - {\left(20\right)}^{2} + 40 \left(20\right) - 16$

$y = - 400 + 800 - 16$

$y = 384$

Therefore, the vertex is at $\left(20 , 384\right)$.

Hope this helps!