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

Let’s factor. \(H(y) = (2y-5)(y-6)\)

\(H\) has two real zeros, \(\answer {\frac {5}{2}}\) and \(\answer {6}\) (in numerical order).

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

The leading coefficient of \(H\) is \(\answer {2}\).

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

\(\lim \limits _{y \to -\infty } H(y) = \answer {\infty }\)

\(\lim \limits _{y \to \infty } H(y) = \answer {\infty }\)

Let’s rewrite \(H\) as a completed square.

\begin{align*} H(y) & = 2y^2 - 17y + 30 \\ & = 2 \left ( y^2 - \frac {17}{2} y \right ) + 30 \\ & = 2 \left ( y^2 - \frac {17}{2} y + \frac {289}{16} - \frac {289}{16} \right ) + 30 \\ & = 2 \left ( y^2 - \frac {17}{2} y + \frac {289}{16} \right ) - \frac {289}{8} + 30 \\ & = 2 \left ( y - \frac {17}{4} \right )^2 - \frac {289}{8} + 30 \\ & = 2 \left ( y - \frac {17}{4} \right )^2 - \frac {49}{8} \end{align*}

Because \(H\) is a quadratic function with a positive leading coefficient, \(H\) decreases on \(\left ( \answer {-\infty }, \answer {\frac {17}{4}} \right ]\) and increases on \(\left [ \answer {\frac {17}{4}}, \answer {\infty } \right )\).

Because \(H\) decreases on \(\left ( -\infty , \frac {17}{4} \right ]\) and increases on \(\left [ \frac {17}{4}, \infty \right )\), \(H\) has a global minimum maximum of \(\answer {-\frac {49}{8}}\) at \(\answer {\frac {17}{4}}\).

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

The only local minimum value of \(H\) is the global minimum of \(\answer {-\frac {49}{8}}\) at \(\answer {\frac {17}{4}}\).

\(H\) is continuous with no maximum value and a minimum value of \(-\frac {49}{8}\). Therefore, the range of \(H\) is