We have,$f(x) = x^{2} + 2x - 5$.
First,find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(x^{2} + 2x - 5) = 2x + 2$.
To find the critical points,set $f'(x) = 0$:
$2x + 2 = 0 \Rightarrow 2x = -2 \Rightarrow x = -1$.
The point $x = -1$ divides the real line into two disjoint intervals: $(-\infty, -1)$ and $(-1, \infty)$.
Case $1$: In the interval $(-\infty, -1)$,choose a test point,say $x = -2$.
$f'(-2) = 2(-2) + 2 = -4 + 2 = -2 < 0$.
Since $f'(x) < 0$ for all $x \in (-\infty, -1)$,the function $f$ is strictly decreasing in $(-\infty, -1)$.
Case $2$: In the interval $(-1, \infty)$,choose a test point,say $x = 0$.
$f'(0) = 2(0) + 2 = 2 > 0$.
Since $f'(x) > 0$ for all $x \in (-1, \infty)$,the function $f$ is strictly increasing in $(-1, \infty)$.
Conclusion: The function is strictly decreasing on $(-\infty, -1)$ and strictly increasing on $(-1, \infty)$.