The last lecture followed the horizontal and the vertical separately. Horizontally x = v_0\cos\theta \, t, vertically y = v_0\sin\theta \, t - \dfrac{1}{2}gt^2. Splitting them turned both into motions we already know well.
The same t sits in both equations. Fix one instant and x is fixed, and so is y. That is how the position of the ball in the figure is decided, and the x and y being measured there always belong to the same instant.
In this form, though, the shape of the path stays hidden. The relation between x and y is only written through t in between. Anyone without a clock cannot read off what line the ball draws.
To learn the shape of the path, then, we get rid of t and rewrite y in terms of x alone.
Getting rid of t does not mean throwing one equation away. We rewrite t in terms of x and put it into the other one.
The horizontal equation is the one to use. Solving x = v_0\cos\theta \, t for t gives t = \dfrac{x}{v_0\cos\theta}. The horizontal speed does not change while the ball is in the air, and it is never zero, so that division always goes through.
The figure carries two rows of marks at equal intervals of time. Along the path the marks bunch up near the top and open out lower down. Along the ground, however, they are evenly spaced everywhere.
That even spacing is what does the work. Time and x are matched one to one, so naming x will do in place of naming t. The ground serves as a ruler, and we can read it instead of a clock.
Put the \dfrac{x}{v_0\cos\theta} just obtained into the t of the vertical equation. Tidying up gives y = x\tan\theta - \dfrac{g}{2v_0^2\cos^2\theta}x^2. No t is left anywhere.
This is a quadratic in x. The coefficient of x^2 is -\dfrac{g}{2v_0^2\cos^2\theta}, and since g, v_0^2 and \cos^2\theta are all positive, the whole of it is negative. A negative leading coefficient means the graph is a parabola opening downward.
You may have noticed that the marks of time have gone from the figure. They are no longer needed. What the equation says is that fixing x fixes y, and no instant appears in that statement.
The two points where the curve meets the ground are marked as well. They are the roots of the quadratic set to y = 0, the launch point and the landing point. The word parabola belongs to this very motion: it is the line drawn by something thrown.
The vertex of a quadratic is the highest point of the parabola. Seen as motion, it is the instant the ball stops rising and turns to falling.
That turning point is where the vertical velocity reaches zero. v_y = v_0\sin\theta - gt vanishes at t = \dfrac{v_0\sin\theta}{g}, and the height at that instant works out to H = \dfrac{v_0^2\sin^2\theta}{2g}. Completing the square gives the same answer, but as physics it is more natural to take it as the moment the vertical velocity disappears.
The easy mistake here is to think the ball stops at the vertex. Only the vertical component has gone to zero; the horizontal component v_0\cos\theta is still there in full. Watch the arrows in the figure. As the ball reaches the top the vertical arrow vanishes and a single horizontal arrow remains. The speed at the highest point is v = v_0\cos\theta.
The parabola is symmetric about the vertical line through its vertex. That the climb takes as long as the descent, and that the speed at a given height is the same going up as coming down, are both restatements of that symmetry.