\(g\) is a constant linear quadratic function, which means its natural domain is all real numbers.

Let’s factor. \(g(t) = (t+1)^2\)

\(g\) has only one real zero, \(\answer {-1}\)

\(g\) is a constant linear quadratic function, which means it is continuous.

The leading coefficient of \(g\) is \(\answer {1}\).

Because \(g\) is a quadratic function with a positive leading coefficient,

\(\lim \limits _{t \to -\infty } g(t) = \answer {\infty }\)

\(\lim \limits _{t \to \infty } g(t) = \answer {\infty }\)

Because \(g\) is a quadratic function with a positive leading coefficient, \(f\) decreases on \(\left ( \answer {-\infty }, \answer {-1} \right ]\) and increases on \(\left [ \answer {-1}, \answer {\infty } \right )\).

Because \(g\) decreases on \(( -\infty , -1 ]\) and increases on \([ -1, \infty )\), \(f\) has a global minimum maximum of \(\answer {0}\) at \(\answer {-1}\).

Because \(\lim \limits _{t \to \infty } g(t) = \answer {\infty }\), \(g\) has no global minimum maximum.

The only local minimum value of \(g\) is the global minimum of \(\answer {0}\) at \(\answer {-1}\).

\(g\) is continuous with no maximum value and a minimum value of \(0\). Therefore, the range of \(g\) is