Đặt \(\sqrt{9-x^2}=t\Rightarrow\left\{{}\begin{matrix}x^2=9-t^2\\0\le t\le3\end{matrix}\right.\)
\(\Rightarrow y=f\left(t\right)=9-t^2+3t=-t^2+3t+9\)
Ta có:
\(-\frac{b}{2a}=\frac{3}{2}\in\left[0;3\right]\)
\(f\left(0\right)=0\) ; \(f\left(\frac{3}{2}\right)=\frac{45}{4}\) ; \(f\left(3\right)=9\)
\(\Rightarrow y_{max}=\frac{45}{4}\) khi \(t=\frac{3}{2}\)
\(y_{min}=9\) khi \(\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)