Consider the function $y(x)=-x^2+6x-3$. It holds that
$$y'(x)=0\Leftrightarrow -2x+6=0\Leftrightarrow x=3.$$
Since $y''(x)=-2<0$ for every $x$, $y(x)$ is a concave function for every $x$. The point $x=3$ is therefore a maximum location of the function $y(x)$. It follows from this that $y(3)=6$ is a maximum of the function $y(x)$.
- $y'(x)=-2x+6$;
- $y''(x)=-2$.
$$y'(x)=0\Leftrightarrow -2x+6=0\Leftrightarrow x=3.$$
Since $y''(x)=-2<0$ for every $x$, $y(x)$ is a concave function for every $x$. The point $x=3$ is therefore a maximum location of the function $y(x)$. It follows from this that $y(3)=6$ is a maximum of the function $y(x)$.