Let \(M(k) = 3(k+2)(k-5)\) with its natural domain.

Domain

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

Zeros

The real zeros of \(M\), in numerical order, are

\[ \answer {-2} \, \text { and } \, \answer {5} \]
Continuity

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

End-Behavior

The leading coefficient of \(M\) is \(\answer {3}\).

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

\(\lim \limits _{k \to -\infty } M(k) = \answer {\infty }\)

\(\lim \limits _{k \to \infty } M(k) = \answer {\infty }\)

Behavior

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 \(\left [ \answer {\frac {3}{2}}, \answer {\infty } \right )\).

Global Extrema

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.

Local Extrema

The only local minimum value of \(M\) is the global minimum of \(\answer {-\frac {147}{4}}\) at \(\answer {\frac {3}{2}}\).

Range

\(M\) is continuous with no maximum value and a minimum value of \(-\frac {147}{4}\). Therefore, the range of \(M\) is

\[ \left [ \answer {-\frac {147}{4}}, \answer {\infty } \right ) \]
2025-01-07 02:34:27