$$\begin{array}{|c|c|}\hline x & h(x) \\ \hline -2 & -2 \\ \hline 1 & 7 \\ \hline 4 & 16 \\ \hline \end{array}$$

The table above shows selected values for the function $h$. In the $xy$-plane, the graph of $y=h(x)$ is a line. What is the value of $h(6)$ ?

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Function notation provides a different way of presenting the standard $xy$ table of values.

$$\begin{array}{|c|c|}\hline x & h(x) \text{ or } y \\ \hline -2 & -2 \\ \hline 1 & 7 \\ \hline 4 & 16 \\ \hline \end{array}$$

We can find the slope, or change in $y$ per change of $x$, as usual. From $(1,7)$ to $(4,16)$,

$$m = \frac{16-7}{4-1}$$ $$m= \frac{9}{3}$$ $$m=3$$

Therefore, if we want to find $h(6)$, we need only extend the table and use the appropriate rate of change.

$$\begin{array}{|c|c|}\hline x & h(x) \text{ or } y \\ \hline -2 & -2 \\ \hline 1 & 7 \\ \hline 4 & 16 \\ \hline 5 & 19 \\ \hline 6 & \boxed{22} \\ \hline \end{array}$$

We can constuct a linear model for the table. Using an arbitrary point on the table and the slope,

$$\begin{array}{|c|c|}\hline x & h(x) \text{ or } y \\ \hline -2 & -2 \\ \hline \colorbox{aqua}{$1$} & \colorbox{aqua}{$7$} \\ \hline 4 & 16 \\ \hline \end{array}$$ $$y-y_1=m(x-x_1)$$ $$y-7=3(x-1)$$

Substituting $x=6$:

$$y-7=3(6-1)$$ $$y-7=15$$ $$y=\boxed{22}$$