Question 1162551
<br>
There are two parabolas that satisfy the given conditions -- one opening up and another opening to the right.<br>
Assuming a parabola that opens up....<br>
The vertex form of the equation of a parabola is<br>
{{{(y-k) = a(x-h)^2}}}<br>
where the vertex is (h,k).<br>
So the equation of this parabola is<br>
{{{(y+4) = a(x+4)^2}}}<br>
Use the coordinates of the known point to find the coefficient a:<br>
{{{(0+4) = a(0+4)^2}}}
{{{4 = 16a}}}
{{{a = 1/4}}}<br>
The equation in vertex form is<br>
{{{(y+4) = (1/4)(x+4)^2}}}<br>
Converting to standard form....<br>
{{{y+4 = (1/4)(x^2+8x+16)}}}
{{{y+4 = (1/4)x^2+2x+4}}}
{{{y = (1/4)x^2+2x}}}<br>
A graph, showing the parabola passing through (0,0)....<br>
{{{graph(400,400,-6,2,-6,2,(1/4)x^2+2x)}}}<br>