Question

Graph a parabola whose vertex is at (3,5) with y-intercept at y= 1.

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\qquadvertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\\rule{31em}{0.25pt}\\\\\stackrel{vertex}{\begin{cases}h=3\\k=5\end{cases}}\implies y=a(x-3)^2+5\qquad \stackrel{y-intercept}{\begin{cases}x=0\\y=1\end{cases}}\implies 1=a(0-3)^2+5\\[2em]-4=a(-3)^2\implies -4=9a\implies -\cfrac{4}{9}=a\\[2em]~~~~~~~~~~~~~~~~y=-\cfrac{4}{9}(x-3)^2+5`