Question 1147599
<pre>{{{drawing(400,2400/11,-2,9,-4,2, graph(400,2400/11,-2,9,-4,2,sqrt(x+1)-1),
graph(400,2400/11,-2,9,-4,2,-sqrt(x+1)-1),
circle(0,0,.1), circle(3,1,.1),
 circle(8,-4,.1),

locate(0,0,"(0,0)"), locate(3,1,"(3,1)"), locate(8,-3.5,"(8,-4)")

 )}}}

Parabolas like this have standard equation:

{{{(y-k^"")^2=4p(x-h^"")}}}

Substitute (x,y) = (0,0):

{{{(0-k)^2=4p(0-h)}}}
{{{k^2=4p(-h)}}}
{{{k^2=-4ph}}}

{{{(y-k^"")^2=4p(x-h^"")}}}

Substitute (x,y) = (3,1):

{{{(1-k)^2=4p(3-h)}}}
{{{1-2k+k^2=12p-4ph)}}}
Substitute +k² for -4ph
{{{1-2k+k^2=12p+k^2)}}}
{{{1-2k=12p}}}
{{{12p+2k=1}}}

Substitute (x,y) = (8,-4):

{{{(-4-k)^2=4p(8-h)}}}
{{{16+8k+k^2=32p-4ph)}}}
Substitute +k² for -4ph
{{{16+8k+k^2=32p+k^2)}}}
{{{16+8k=32p}}}
{{{32p-8k=16}}}
That can be divided through by 8
{{{4p-k=2}}}

Solve this system:

{{{system(12p+2k=1,4p-k=2)}}}

And you'll get p=1/4, k=-1

Substitute in {{{k^2=-4ph}}}

{{{(-1)^2=-4(1/4)h}}}
{{{1=-h}}}
{{{h=-1}}}

So substitute for h,k, and p:

{{{(y-(-1)^"")^2=4(1/4)(x-(-1)^"")}}}

{{{(y+1^"")^2=(1)(x+1^"")}}}

{{{(y+1^"")^2=x+1^""}}}    <---answer

Edwin</pre>