Question 621020
For hyperbolas, we like the standard form of the equation, the one that shows a difference of squares equal to 1. 
It may look like
{{{x^2/a^2-y^2/b^2=1}}} or {{{y^2/a^2-x^2/b^2=1}}} .
That form of the equation shows you all the numbers you need to know to figure out the he foci, vertices and the asymptotes. (All you need is the a and b numbers).
{{{y^2-x^2/15=1}}} is the equation (in standard form) of a hyperbola centered at the origin, so this is an easy problem.
We know this one is centered at the origin because there is just an {{{x^2}}} and a {{{y^2}}}, with nothing added or subtracted before squaring.
Because of that simplicity, it is easy to see that changing x to -x gives you the same equation, meaning that the graph is symmetrical with respect to the y-axis. The same can be said of changing y to -y, and the symmetry with respect to the x-axis.

For y=0 we would have a negative number equal to 1 {{{-x^2/15=1}}} and that cannot be.
So, we can see that the graph does not touch the x-axis, where y=0. (In fact the graph does not even want to get close to the x-axis)
On the other hand, y cannot be zero, but x can be zero.
When {{{x=0}}}, you see that {{{y^2=1}}} , meaning {{{y=1}}} or {{{y=-1}}} , so the graph goes through the points (0,1) and (0,-1).
For all other points, {{{x^2>0}}} so {{{x^2/15>0}}} and {{{y^2=x^2/15+1>1}}} , meaning that all the other points are even farther away from the x-axis, where y=0.
The closest that the hyperbola comes to the x-axis is the points (0,1) and (0,-1) , which are the vertices.
As x (and y) grow larger in absolute value, {{{x^2}}} and {{{y^2}}} grow larger, and the graph gets closer to the asymptotes.
A little algebra transforms the equation into one that gives us the equations of the asymptotes:
{{{y^2-x^2/15=1}}} --> {{{y^2+1=x^2/15}}} --> {{{(y^2+1)/x^2=1/15}}} --> {{{y^2/x^2+1/x^2=1/15}}} --> {{{y^2/x^2=1/15+1/x^2}}}
As {{{x^2}}} grows larger, {{{1/x^2}}} grows smaller, and the graph gets closer to the graph for
{{{y^2/x^2=1/15}}} which is the graph for the lines
{{{y/x=sqrt(1/15)}}} <--> {{{y=sqrt(1/15)x}}} and
{{{y/x=-sqrt(1/15)}}} <--> {{{y=-sqrt(1/15)x}}} .
Those lines are the asymptotes.
Because teachers do not like to see square roots in denominators, we may have to write them as 
{{{y=(sqrt(15)/15)x}}} and {{{y=-(sqrt(15)/15)x}}} .
THE FOCI:
There foci are at a distance {{{c}}} from the center of the hyperbola, and the number {{{c}}} is related to the numbers {{{a^2}}} and {{{b^2}}} in the standard form of the equation by a formula that can be derived using the Pythagorean theorem. It is
{{{c^2=a^2+b^2}}}
In this case, your {{{a^2}}} and {{{b^2}}} are 1 and 15, so
{{{c^2=1+15}}} --> {{{c^2=16}}} --> {{{c=4}}}.
The center was (0,0) (the origin).
The vertices were ((0,-1) and (0,1), on the y-axis.
The foci are on the same line, but at distance 4 from the center/origin, at
(0,-4) and (0,4).