[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] ~\dotfill\\\\ f(x)=-2(x+4)^2-3\implies f(x)=-2[x-(\stackrel{h}{-4})]^2\stackrel{k}{-3}~\hfill \stackrel{vertex}{(-4,-3)}[/tex]
well, to get a second point, we simply pick a random "x" hmmm say x = 1, so then
f( 1 ) = -2( 1 + 4) ² - 3
f( 1 ) = -2(25) - 3
f( 1 ) = -53
so from that we get the point of x = 1, y = -53 ( 1 , -53), so it looks like the picture below.
