[tex]\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ y = -6x^2-1\implies y = -6(x-\stackrel{h}{0})^2\stackrel{k}{-1}\qquad \qquad \stackrel{vertex}{(0,-1)}\qquad \qquad \stackrel{maximum}{y = -1}[/tex]
Answer:
maximum = -1
Step-by-step explanation:
when graphing, it is an upside down parabola, with the vertex [and maximum] at -1
=> look at the graph
:)
