## Precalculus (6th Edition)

$-1$
The given equation can be written as: $f(x)=(x^2+2x+1)+3 \\f(x)=(x+1)^2+3 \\f(x) = [x-(-1)]^2+3$ This means the given equation is equivalent to $f(x) = [x-(-1)]^2+3$. RECALL: The vertex form of a quadratic function is $f(x)=a(x-h)^2+k$ where $(h, k)$ is the vertex. The function $f(x) = [x-(-1)]^2+3$ is in vertex form. The vertex is $(-1, 3)$. Thus, the missing expression in the given statement is: $-1$