105c8f…: “Subsequent to the navy equation”

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

Describe the graph of $y = M(k)$.

Shape

$M$ is a constant linear quadratic function, which means its graph is a line parabola.

The leading coefficient of $M$ is $\answer {3}$. Therefore, the parabola opens up down

Intercepts

The real zeros of $M$ are $-2$ and $5$, which make the intercepts

\[ \left ( \answer {-2}, \answer {0}) \, \text { and } \, (\answer {5}, \answer {0} \right ) \]

Breaks

Because the graph is a parabola, it has no breaks.

Axis of Symmetry

The midpoint of $-2$ and $5$ is $\answer {\frac {3}{2}}$, therefore, the axis of symmetry has the equation

\[ k = \answer {\frac {3}{2}} \]

Vertex

\[ M\left ( \frac {3}{2} \right ) = -\frac {147}{4} \]

Therefore, the vertex is

\[ \left ( \answer {\frac {3}{2}}, \answer {-\frac {147}{4}} \right ) \]

2025-01-07 02:38:23