Question

F(x)=(x+1)^2-8 Give axis of symmetry and vertex

Answer 1

$\bf ~~~~~~\textit{parabola vertex form} \\\\ y=a(x- h)^2+ k~\hspace{7em}vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=(x+1)^2-8\implies y=1[x-(\stackrel{h}{-1})]^2+(\stackrel{k}{-8})~\hfill \stackrel{vertex}{(-1,-8)}$

the equation is in x-terms, meaning is a vertical parabola, and therefore the axis of symmetry will be the x-coordinate of the vertex, namely x = -1.