Vertex form is f(x)=a(x-p)^2 +q. how do i determine a?

Y1 is the simplest parabola.  Its vertex is at (0,0) and it passes thru (2,4).  This is enough info to conclude that y1 = x^2.

y4, the lower red graph, is a bit more of a challenge.  We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).

Let's try this:  assume that the general equation for a parabola is 
y-k = a(x-h)^2, where (h,k) is the vertex.  Subst. the known values,

-3-(-4) = a(2-0)^2.  Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.

The equation of parabola y4 is   y+4 = (1/4)x^2

Or you could elim. the fraction and write the eqn as 4y+16=x^2, or

4y = x^2-16, or    y = (1/4)x - 4.  Take your pick!  Hope this helps you find "a" for the other parabolas.

Def is a scalene triangle