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

\(p(x)\) doesn’t factor easily. factor. We’ll use the quadratic formula.

\(2x^2 - x + 3 = \)

\(a = \answer {2}\)

\(b = \answer {-1}\)

\(c = \answer {3}\)

\(p\) has no real zeros.

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

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

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

\(\lim \limits _{x \to -\infty } p(x) = \answer {\infty }\)

\(\lim \limits _{x \to \infty } p(x) = \answer {\infty }\)

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

\begin{align*} p(x) & = 2x^2 - x + 3 \\ & = 2 \left ( x^2 - \frac {1}{2} x \right ) + 3 \\ & = 2 \left ( x^2 - \frac {1}{2} x + \frac {1}{16} - \frac {1}{16} \right ) + 3 \\ & = 2 \left ( x - \frac {1}{4} \right )^2 - \frac {1}{16} + 3 \\ & = 2 \left ( x - \frac {1}{4} \right )^2 + \frac {47}{16} + 3 \end{align*}

Because \(p\) is a quadratic function with a positive leading coefficient, \(p\) decreases on \(\left ( \answer {-\infty }, \answer {\frac {1}{4}} \right ]\) and increases on \(\left [ \answer {\frac {1}{4}}, \answer {\infty } \right )\).

Because \(p\) decreases on \(\left ( -\infty , \frac {1}{4} \right ]\) and increases on \(\left [ \frac {1}{4}, \infty \right )\), \(p\) has a global minimum maximum of \(\answer {\frac {47}{16}}\) at \(\answer {\frac {1}{4}}\).

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

The only local minimum value of \(p\) is the global minimum of \(\answer {\frac {47}{16}}\) at \(\answer {\frac {1}{4}}\).

\(p\) is continuous with no maximum value and a minimum value of \(\frac {47}{16}\). Therefore, the range of \(p\) is