Function [tex]k[/tex] is a continuous quadratic function that includes the ordered pairs shown in the table.

[tex]\[
\begin{array}{c|ccccc}
\text{x} & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline
k(x) & 5 & 8 & 9 & 8 & 5 & 0 \\
\end{array}
\][/tex]

Over which interval of the domain is the function increasing?

A. [tex](1, \infty)[/tex]
B. [tex](-\infty, 1)[/tex]
C. [tex](-\infty, \infty)[/tex]
D. [tex](-\infty, 9)[/tex]



Answer :

To determine the interval over which the function [tex]\( k \)[/tex] is increasing, we need to analyze the nature of the quadratic function [tex]\( k(x) = ax^2 + bx + c \)[/tex] given by the ordered pairs:

[tex]\[ \begin{array}{c|cccccc} x & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline k(x) & 5 & 8 & 9 & 8 & 5 & 0 \\ \end{array} \][/tex]

From these points, we can see that the function [tex]\( k(x) \)[/tex] has the general form [tex]\( k(x) = ax^2 + bx + c \)[/tex].

To understand when the quadratic function is increasing, we'll examine its vertex. The vertex of a quadratic function [tex]\( k(x) = ax^2 + bx + c \)[/tex] occurs at [tex]\( x = -\frac{b}{2a} \)[/tex]. This vertex represents a maximum or minimum point of the parabola.

Given the form of the quadratic equation and observing the pattern of the values from the table, we can deduce the following:
- The function [tex]\( k(x) \)[/tex] reaches its maximum at [tex]\( x = 1 \)[/tex].
- Quadratic functions with a maximum vertex are increasing to the right of the vertex (i.e., for [tex]\( x > 1 \)[/tex]).

Since the function is symmetric around its vertex and increasing to the right of this point, it will be increasing for all [tex]\( x > 1 \)[/tex].

Thus, the interval over which the function [tex]\( k \)[/tex] is increasing is:

[tex]\[ \boxed{D. (-\infty, 9)} \][/tex]