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

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 \(h(6)\) ?

Summary Submit Skip Question

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 \(x=6\):

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