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

Mar 16, 2017

$\left(- 1 , 4\right)$

Explanation:

There is a lovely and straightforward (which makes it all the lovelier) rule for working out vertices such as this one.

Think of the general parabola: $y = a {x}^{2} + b x + c$, where $a \ne 0$

The formula for finding the $x$-vertex is $\frac{- b}{2 a}$ and to find the $y$-vertex, you insert the value you found for $x$ into the formula.

Using your question $y = - {x}^{2} - 2 x + 3$ we can establish the values of $a , b ,$and $c$.

In this case:

$a = - 1$
$b = - 2$; and
$c = 3$.

To find the $x$-vertex we need to replace the values for $a$ and $b$ in the formula given above ($\textcolor{red}{\frac{- b}{2 a}}$):

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

So we now know that the $x$-vertex is at $- 1$.

To find the $y$-vertex, go back to the original question and replace all the instances of $x$ with $- 1$:

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

$y = - {\left(- 1\right)}^{2} - 2 \cdot \left(- 1\right) + 3$

$y = - 1 + 2 + 3$

$y = 4$

We now know that the $x$-vertex is at $- 1$ and the $y$-vertex is at $4$ and this can be written in coordinate format:

$\left(- 1 , 4\right)$