Question 1087418
 Find the standard form {{{f(x)=a(x-h)^2+k}}} of the quadratic function that has a vertex at

 ({{{-1/4}}},{{{-1/6}}}) =>{{{h=-1/4}}}  and {{{k=-1/6}}}

so far, equation is:

{{{f(x)=a(x-(-1/4))^2+(-1/6)}}}

{{{f(x)=a(x+1/4)^2-1/6}}}


and if x-intercept of {{{-3/4 }}}=>  {{{x=-3/4 }}} when  {{{f(x)=0 }}}

use it to find {{{a}}}:

{{{0=a(-3/4+1/4)^2-1/6}}}

{{{0=a(-2/4)^2-1/6}}}

{{{1/6=a(-1/2)^2}}}

{{{1/6=a(1/4)}}}

{{{a=(1/6)/(1/4)}}}

{{{a=(4/6)}}}

{{{a=(2/3)}}}

so, your equation is:

{{{f(x)=(2/3)(x+1/4)^2-1/6}}}


{{{drawing( 600, 600, -6, 6, -6, 6, 
circle(-1/4,-1/6,.06),locate(-1/4,-1/6,V(-1/4,-1/6)),
circle(-3/4,0,.07),locate(-1.3,0.6,p(-3/4,0)),
 graph( 600, 600, -6, 6, -6, 6, (2/3)(x+1/4)^2-1/6)) }}}