Vertex form is #y=a(x - h)^2 + k#, where #(h, k)# is the vertex. From that information, our equation should look like this:
#y=(x-1)^2-1#
That's part of the solution, but we still need to include the #y#-intercept somehow. We've filled in all the variables, except for #a#. Let's see what that does on the parent graph, #y=2(x)^2# vs #y=x^2#
graph{y=2x^2}
graph{y=x^2}
It looks like the vertex didn't change, but the graph became narrower. And on the first graph, #x=-1, y=2#, but on the second one, when #x=-1, y=1#. So #a# influences the graph, and could be used to find the #y#-intercept
So, if we want our #y#-intercept to be #3#, we should set #a=3#. WAIT!! Our equation #y=(x-1)^2-1# shows that the graph is shifted down #1# unit. So if we want the #y#-intercept to be #3#, we need to move it up one unit. So our equation will really be #y=4(x+1)^2-1#
Just to double check our work, let's graph it
graph{y=4(x-1)^2-1}
We were right! Good job
