To determine the horizontal intercepts, we'll set \(y=0\) and solve for \(x\text{:}\)

\begin{align*} 0\amp=-2(x+5)^2+12\\ \end{align*}

We cannot immediately factor this equation, but we can use the square root property to solve it. To do so, we'll first need to subtract \(12\) from each side of the equation and then divide each side by \(-2\text{:}\)

\begin{align*} -12\amp=-2(x+5)^2\\ 6\amp=(x+5)^2\\ \end{align*}

Now applying the square root property,

\begin{align*} \pm\sqrt{6}\amp=x+5\\ x\amp=-\sqrt{6}-5,\ x=\sqrt{6}-5\\ x\amp\approx -7.449,\ x\approx -2.551 \end{align*}

The \(x\)-intercepts occur at \((-\sqrt{6}-5,0)\) and \((\sqrt{6}-5,0)\text{,}\) or at approximately \((-7.449,0)\) and \((-2.551,0)\text{.}\)

To determine the \(y\)-intercept, we'll replace \(x\) with \(0\text{:}\)

\begin{align*} y\amp=-2(\substitute{0}+5)^2+12\\ \amp=-2(25)+12\\ \amp=-38 \end{align*}

The \(y\)-intercept occurs at \((0,-38)\text{.}\)

To determine the vertex using the formula \(x=-\frac{b}{2a}\text{,}\) we'll first need to expand this equation so that \(a\) and \(b\) can be identified:

\begin{align*} y\amp=-2(x+5)^2+12\\ y\amp=-2(x^2+10x+25)+12\\ y\amp=-2x^2-20x-50+12\\ y\amp=-2x^2-20x-38 \end{align*}

Now we can use \(a=-2\) and \(b=-20\) to determine the \(x\)-value of the vertex:

\begin{align*} x\amp=-\frac{-20}{2(-2)}\\ \amp=-5 \end{align*}

Lastly, we'll use this value to find the \(y\)-value. We can use either the original equation or the expanded one; we'll use the original one here:

\begin{align*} y\amp=-2(\substitute{-5}+5)^2+12\\ \amp=-2(0)+12\\ \amp=12 \end{align*}

Therefore the vertex is the point \((-5,12)\text{.}\)