Which statement describes the inverse of $m(x)=x^2-17x$?
AThe domain restriction $x \geq \frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}-\sqrt{x+\frac{289}{4}}$.
BThe domain restriction $x \geq \frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}+\sqrt{x+\frac{289}{4}}$.
CThe domain restriction $x \geq-\frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}-\sqrt{x+\frac{289}{4}}$.
DThe domain restriction $x \geq-\frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}+\sqrt{x+\frac{289}{4}}$.
AThe domain restriction $x \geq \frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}-\sqrt{x+\frac{289}{4}}$.
BThe domain restriction $x \geq \frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}+\sqrt{x+\frac{289}{4}}$.
CThe domain restriction $x \geq-\frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}-\sqrt{x+\frac{289}{4}}$.
DThe domain restriction $x \geq-\frac{17}{2}$ results in $m^{-1}(x)=\frac{17}{2}+\sqrt{x+\frac{289}{4}}$.