We determine where the function $y(x)=x^2+5x+7$ is increasing/decreasing.
$y'(x)=2x+5$. Hence, $y'(x)=0$ for $x=-2\frac{1}{2}$. By the use of a sign chart we find the following.
$y'(x)\geq 0$ for $x\geq -2\frac{1}{2}$ and hence, $y(x)$ is increasing for $x\geq -2\frac{1}{2}$.
$y'(x)\leq 0$ for $x\leq -2\frac{1}{2}$ and hence, $y(x)$ is decreasing for $x\leq -2\frac{1}{2}$.
$y'(x)=2x+5$. Hence, $y'(x)=0$ for $x=-2\frac{1}{2}$. By the use of a sign chart we find the following.
$y'(x)\geq 0$ for $x\geq -2\frac{1}{2}$ and hence, $y(x)$ is increasing for $x\geq -2\frac{1}{2}$.
$y'(x)\leq 0$ for $x\leq -2\frac{1}{2}$ and hence, $y(x)$ is decreasing for $x\leq -2\frac{1}{2}$.