Question 1176619


the quadratic function in vertex form, some refer to it as  “standard form”

{{{f (x) = a(x - h)^2 + k}}}, where ({{{h}}}, {{{k}}}) is the vertex of the parabola 

if given  a vertex ​(​{{{2}}},​{{{1}}}) then {{{h=2}}} and{{{k=1}}}

 {{{f (x) = a(x - 2)^2 + 1}}}



if passes through the point ​({{{3}}}​,{{{5}}}) we have

{{{5 = a(3 -2)^2 + 1}}}
{{{5 = a+1 }}}
{{{a=5-1}}}
{{{a=4}}}

and your equation is:

{{{f (x) = 4(x -2)^2 + 1}}}

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,1,.12),locate(2,1,V(2,1)),
circle(3,5,.12),locate(3,5,p(3,5)),
 graph( 600, 600, -10, 10, -10, 10, 4(x -2)^2 + 1)) }}}