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y=x^2+1 find the vertex
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\[\Huge \frac{-b}{2a}\]
y=x^2+1 really looks like y=1(x-0)^2+1 and it is in the form y = a(x-h)^2+k where (h,k) is the vertex We can see that h = 0 and k = 1 So the vertex is (0, 1)
y = x^2 + 1 is just y = x^2 but shifted vertically by +1. So the vertex of y = x^2 is just (0,0) and so the vertex of y = x^2 + 1 is (0,1). :)
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