Let \(M(k) = -3(k+2)^2 - 5\) be a quadratic function.

Since \(M(k)\) is a quadratic function, its domain is \(\left ( \answer {-\infty }, \answer {\infty } \right )\).

Since \(M(k)\) is a quadratic function, it is continuous and has no discontinuities or singularities.

Zeros

In numerical order, the zeros of \(M(k)\) are

\[ \answer {-2 - \sqrt {\frac {5}{3}}} \, \text { and } \, \answer {-2 + \sqrt {\frac {5}{3}}} \]
End-Behavior

Since \(M(k)\) is a quadratic function with a negative leading coefficient,

\[ \lim \limits _{k \to -\infty }M(k) = \answer {-\infty } \, \text { and } \, \lim \limits _{k \to \infty }M(k) = \answer {-\infty } \]

This means \(M(k)\) has no global maximum minimum

Rate of Change

\(iRoC_M(k) = \answer {-6(k+2)}\).

\(iRoC_M(k) = 0\) at \(k = \answer {-2}\).

\(\answer {-2}\) is the only critical number of \(M(k)\).

\(iRoC_M(k) > 0\) on \(\left ( \answer {-\infty }, \answer {-2} \right )\).

\(iRoC_M(k) < 0\) on \(\left ( \answer {-2}, \answer {\infty } \right )\).

Behavior

\(M(k)\) increases on \(\left ( \answer {-\infty }, \answer {-2} \right )\).

\(M(k)\) decreases on \(\left ( \answer {-2}, \answer {\infty } \right )\).

The maximum of \(M(k)\) is located at \(k = \answer {-2}\).

The maximum of \(M(k)\) is \(\answer {-5}\).

This global maximum is also a local maximum.

Range

The range of \(M(k)\) is

\[ \left ( \answer {-\infty }, \answer {-5} \right ] \]
2025-05-18 05:24:59