Question 924934
Quadratic curve : {{{y=ax^2+bx+c}}}
Passes through the origin : 
{{{0=a(0)+b(0)+c}}}
{{{c=0}}}
So,
{{{y=ax^2+bx}}}
The derivative of the function is,
{{{dy/dx=2ax+b}}}
The function also passes through (-2,3) where, 
{{{dy/dx=-1}}}
{{{2a(-2)+b=-1}}}
1.{{{-4a+b=-1}}}
But also from the value of the function,
{{{3=a(-2)^2+b(-2)}}}
2.{{{4a-2b=3}}}
Add eq. 1 to eq. 2,
{{{-4a+b+4a-2b=-1+3}}}
{{{-b=2}}}
{{{b=-2}}}
Then,
{{{-4a-2=-1}}}
{{{-4a=1}}}
{{{a=-1/4}}}
So then,
{{{highlight(y=-(1/4)x^2-2x)}}}