Question

How do you find the vertex of `y= -x^2 -3`?

Answer 1

So we know that in the equation `y=a(x-d)^2+e`, the vertex is `(d,e)` because it's a parabola. We know that `a=-1` because the the only way to get `-x^2` is by doing `ax^2`, so `ax^2=-x^2`, so `a = -1`.

We also have that `-2adx=0x` because there is no `x` factor and the only way to get something with only one `x` in it is by using `-2adx`. If `-2adx=0x`, we can divide both sides by `-2x` to get `ad=0`. For this to be true, then either `a` or `d` has to be `0`, but we already know that `a` is `-1`, so `d=0`.

Now the only way to get a number with no `x` in it is `ad^2+e`, so `ad^2+e=-3`. We know that `a` and `d` are `-1` and `0`, respectively, so we replace the variables to get `e=-3`. We know that the vertex is `(d,e)`, so we replace the variables to get that the vertex is `(0,-3)`.