Question

Where is tge vertex of the graph of y=-x^2

Answer 1

`\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\ -------------------------------\\\\ y=-x^2\implies y=-1(x-\stackrel{h}{0})^2+\stackrel{k}{0}`