Answer
The equation matches with the graph labeled as VI.
Work Step by Step
It has the equation of a parabola with horizontal axis: $y^2=4px$
$y^2-x=0$
$y^2=8x$
$4p=8$
$p=2$
$p\gt0$. Parabola opens to the right.
When $x=1$
$y^2=8(1)=8$
$y=±2\sqrt 2\approx±2.83$
So, the parabola passes through the points $(1,±2.83)$
The equation matches with the graph labeled as VI.