# How do you find the vertex of y= -x^2 -3?

So we know that in the equation $y = a {\left(x - d\right)}^{2} + e$, the vertex is $\left(d , e\right)$ because it's a parabola. We know that $a = - 1$ because the the only way to get $- {x}^{2}$ is by doing $a {x}^{2}$, so $a {x}^{2} = - {x}^{2}$, so $a = - 1$.
We also have that $- 2 a \mathrm{dx} = 0 x$ because there is no $x$ factor and the only way to get something with only one $x$ in it is by using $- 2 a \mathrm{dx}$. If $- 2 a \mathrm{dx} = 0 x$, we can divide both sides by $- 2 x$ to get $a d = 0$. For this to be true, then either $a$ or $d$ has to be $0$, but we already know that $a$ is $- 1$, so $d = 0$.
Now the only way to get a number with no $x$ in it is $a {d}^{2} + e$, so $a {d}^{2} + e = - 3$. We know that $a$ and $d$ are $- 1$ and $0$, respectively, so we replace the variables to get $e = - 3$. We know that the vertex is $\left(d , e\right)$, so we replace the variables to get that the vertex is $\left(0 , - 3\right)$.