We determine the stationary points of the function $y(x)=x^2+5x+7$.
$y'(x)=2x+5$. Hence, $y'(x)=0$ if $x=-2\frac{1}{2}$.
Consequently, the only stationary point is $x=-2\frac{1}{2}$.
$y'(x)=2x+5$. Hence, $y'(x)=0$ if $x=-2\frac{1}{2}$.
Consequently, the only stationary point is $x=-2\frac{1}{2}$.