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) and x-intercept of -3/4
<pre>{{{f(x) = y = a(x - h)^2 + k}}}
{{{0 = a(- 3/4 - - 1/4)^2 - 1/6}}} ------- Substituting {{{matrix(1,16, "(", - 3/4, ",", "0)", for, "(x", "y)", and, "(", - 1/4, ",", - 1/6, ")", for, "(h,", "k)")}}}
{{{0 = a(- 3/4 + 1/4)^2 - 1/6}}}
{{{1/6 = a(- 1/2)^2}}}
{{{1/6 = (1/4)a}}}
{{{a = (1/6)/(1/4)}}}
{{{matrix(1,5, a, "=", (1/6) * 4, "=", 2/3)}}}
{{{highlight_green(f(x) = (2/3)(x + 1/4)^2 - 1/6)}}} ------- Substituting {{{matrix(1,10, 2/3, for, "a,", - 1/4, for, "h,", and, - 1/6, for, k)}}}