Y=x^2 -x -6 quadratic function

y=x^2 -x -6 this is the function. The exercise is to find the culmination points, zero points, cutting point with the ordinate axis and plot a graph.
I understand these:
- Culmination point, you use the formula -b/2a
- Zero point, where the parabola cuts with x axis(on both - and + sides I guess?)
- The cutting point with ordinate axis <--- I'm not so sure about this one

I have done all the previous things, but what exactly am I asked with the ordinate axis? The result for this should be (0;-6)

Thanks in advance!

That's just asking for the y-intercept in another way. Where does it intersect the y axis? Unlike the x-intercepts (zero points), there is only one, at most, for any function.

Set x to 0 and solve for y:

y = x^2 - x - 6 = 0^2 - 0 - 6 = -6
(0, -6)

It's when they cross the x and y axes. crossed the x axis, when y = 0. This means you factorise into (x-3)(x+2)=0
x = 3&-2

crossed the y axis, when x = 0
put 0 into the formula
0-0-6=-6