Question 177930
{{{(1/k)(y-k)=(x-h)^2}}}...vertex is(-3,4)
:
{{{(1/k)(y-4)=(x+3)^2}}} and we have another point(-6,0)
:
{{{(1/k)(0-4)=(-6+3)^2}}}
:
{{{(1/k)(-4)=9}}}
:
{{{(1/k)=-9/4}}}
:
{{{(-9/4)(y-4)=(x+3)^2}}}