What is the the vertex of #y =x^2-6x-7 #?

There are different ways this can be done.

This equation is in standard form, so you can use the formula #P(h,k) = (-b/(2a),-d/(4a))# Where the (d) is the discriminant. #d = b^2-4ac#

Or to save time, you can find the (x) coordinat for the vertex with #-b/(2a)# and put the result back in to find the (y) coordinat.

Alternatively, you can rearrenge the equation into vertex form:
#a(x-h)^2+k#

To do this start by putting a outside the brackets. This is easy because #a=1#

#x^2-6x-7 = 1(x^2-6x) - 7#

Now we have to change #x^2-6x# into #(x-h)^2#
To do this we can use the quadratic sentence: #(q-p)^2 = q^2+p^2-2qp#

Let's say #q=x# therefore we get:
#(x-p)^2 = x^2+p^2-2xp#

This looks sort of what we need, but we are still far of, as we only have #x^2#.

If we look at #x^2-6x#, we can se that there is only one part raised to the power of two, therefore #p^2# must be removed. This means:

#(x-p)^2-p^2=x^2-2xp#

Looking at the right side, we can see it is almost #x^2-6x#, in fact we only have to solve #-2xp = -6x# #iff p = 3#

This means:
#(x-3)^2-9 = x^2-6x#

Another way of doing it would be to make a qualified guess and use the quadratic sentences to see if it is correct.

Now go back to our original formula and replace #x^2-6x# with #(x-3)^2-9#

We get:

#1(x^2-6x) - 7 = 1((x - 3)^2-9) - 7 = 1(x - 3)^2-9 - 7 = 1(x - 3)^2-16#
This is similar to the vertex form:
#a(x-h)^2+k#
Where
#h = 3# and #k=-16#

When the quadratic equation is in vertex form, the vertex is simply the point #P(h,k)#

Therefore the vertex is #P(3,-16)#