# What is the vertex of y=-x^2-3x+9 ?

Jun 26, 2018

Vertex: $\left(- 1.5 , 11.25\right)$

#### Explanation:

$y = - {x}^{2} - 3 x + 9$

To find the $x$-coordinate of the vertex of a standard quadratic equation ($y = a {x}^{2} + b x + c$), we use the formula $\frac{- b}{2 a}$.

We know that $a = - 1$ and $b = - 3$, so let's plug them into the formula:
$x = \frac{- \left(- 3\right)}{2 \left(- 1\right)} = \frac{3}{-} 2 = - 1.5$

To find the $y$-coordinate of the vertex, just plug in the $x$-coordinate back into the original equation:
$y = - {\left(- 1.5\right)}^{2} - 3 \left(- 1.5\right) + 9$

$y = - 2.25 + 4.5 + 9$

$y = 11.25$

Therefore, the vertex is at $\left(- 1.5 , 11.25\right)$.

Here's a graph of this equation (desmos.com):

As you can see, the vertex is indeed at $\left(- 1.5 , 11.25\right)$.

Hope this helps!