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The correct answer is $-7$. For a quadratic function defined by an equation of the form  $f\left(x\right)=a{\left(x-h\right)}^2+k$, where $a$, $h$, and $k$ are constants and $a>0$, the function reaches its minimum when $x=h$. In the given function, $a=1$, $h=-7$, and $k=4$. Therefore, the value of $x$ for which $f\left(x\right)$ reaches its minimum is $-7$.