Tracing of the Parabola $y^2=4ax$

Once we know the equation is $y^2=4ax$ (with $a>0$), we can work out the shape of the curve purely by studying the algebra, without plotting many individual points. Four observations, taken together, pin down the shape completely.

1. Symmetry. Solving for $y$ gives $y=\pm2\sqrt{ax}$. So for every valid value of $x$, there are two values of $y$ that are negatives of each other. This means the curve is symmetric about the $X$-axis — whatever the curve does above the axis, it mirrors exactly below the axis.

2. Region covered. For $x<0$, the quantity $ax$ is negative (since $a>0$), so $y=\pm2\sqrt{ax}$ is not a real number. This means the entire curve lies to the right of the $Y$-axis — there is no part of the parabola with a negative $x$-coordinate.

3. Intersection with the axes. Setting $x=0$ gives $y^2=0$, i.e. $y=0$. So the curve meets the coordinate axes at exactly one point, the origin $O(0,0)$ — this is, of course, the vertex.

4. Behaviour as $x$ grows. As $x\to\infty$, $y\to\pm\infty$ as well. So the curve keeps extending outward without bound, always opening further to the right, never closing back up on itself (unlike an ellipse). …