Question

Identify the vertex and the y-intercept of the graph of the function,
y = 0.25(x + 5)2

Answer 1

`\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] \rule{34em}{0.25pt}\\\\ y=0.25(x+5)^2\implies y=0.25[x-(\stackrel{h}{-5})]^2+\stackrel{k}{0}~\hfill \stackrel{\textit{vertex}}{(-5,0)} \\\\[-0.35em] ~\dotfill`

`\bf \stackrel{\textit{the y-intercept occurs when x =0}~\hfill }{y=0.25(0+5)^2\implies y=0.25(5)^2}\implies y=6.25~\hfill \stackrel{\textit{y-intercept}}{(0, 6.25)}`