Let \(M(k) = 3(k+2)(k-5)\) with its natural domain.
\(M\) is a constant linear quadratic function, which means its natural domain is all real numbers.
The real zeros of \(M\), in numerical order, are
The leading coefficient of \(M\) is \(\answer {3}\).
Because \(M\) is a quadratic function with a positive leading coefficient,
\(\lim \limits _{k \to -\infty } M(k) = \answer {\infty }\)
\(\lim \limits _{k \to \infty } M(k) = \answer {\infty }\)
Because \(M\) is a quadratic function with a positive leading coefficient, \(M\) decreases on \(\left ( \answer {-\infty }, \answer {\frac {3}{2}} \right ]\) and increases on \(\left [ \answer {\frac {3}{2}}, \answer {\infty } \right )\).
Because \(M\) decreases on \(\left ( -\infty , \frac {3}{2} \right ]\) and increases on \(\left [ \frac {3}{2}, \infty \right )\), \(M\) has a global minimum maximum of \(\answer {-\frac {147}{4}}\) at \(\answer {\frac {3}{2}}\).
Because \(\lim \limits _{k \to \infty } M(k) = \answer {\infty }\), \(M\) has no global minimum maximum.