Question 1125823
  

Find the quadratic function that has given vertex and goes through the given point. 

vertex: ({{{5}}},{{{6}}}), 
point: ({{{-1}}},{{{0}}}) 


first, we recall a function {{{f(x)=a(x-h)^2+k}}}, which will give us vertex ({{{h}}},{{{k}}})=({{{5}}},{{{6}}}) 

So the function we are looking for should look like 
{{{f(x)=a(x-5)^2+6}}}, 

and since the parabola passes through ({{{-1}}},{{{0}}})  , we have 

{{{0=a(-1-5)^2+6}}}, which means 

{{{0=36a+6 }}}

{{{36a=-6}}}

{{{a=-6/36}}}

{{{ a=-1/6}}}

 and equation is:

{{{f(x) = -(1/6)(x-5)^2+6}}}


{{{drawing( 600, 600, -10, 15, -10, 10,
circle(5,6,.12),locate(5,6,V(5,6)),
circle(-1,0,.16),locate(-1.5,0.5,p(-1,0)),
 graph( 600, 600, -10, 15, -10, 10,-(1/6)(x-5)^2+6)) }}}