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

1 Answer
Apr 27, 2018

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)#.