Question

The table below shows some values of a linear function, h(x). Complete the table.
x h(x)
0 45.3
1 54.6
2
3
4

Answer 1

To complete the table for the linear function \( h(x) \), we first need to find the slope of the function using the given values. We can take the points \( (0, 45.3) \) and \( (1, 54.6) \) to find the slope.

The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Using the points:
- \( (0, 45.3) \) where \( y_1 = 45.3 \) and \( x_1 = 0 \)
- \( (1, 54.6) \) where \( y_2 = 54.6 \) and \( x_2 = 1 \)

Substituting into the slope formula:
\[
m = \frac{54.6 - 45.3}{1 - 0} = \frac{9.3}{1} = 9.3
\]

Now that we have the slope \( m = 9.3 \), we can use this to find the values of \( h(x) \) for \( x = 2, 3, \) and \( 4 \).

Using the point-slope form of the equation of a line, the general formula for \( h(x) \) can be given as:
\[
h(x) = h(0) + m \cdot x
\]
Substituting in the known values:
\[
h(x) = 45.3 + 9.3x
\]

Now we can calculate the missing values:

For \( x = 2 \):
\[
h(2) = 45.3 + 9.3 \times 2 = 45.3 + 18.6 = 63.9
\]

For \( x = 3 \):
\[
h(3) = 45.3 + 9.3 \times 3 = 45.3 + 27.9 = 73.2
\]

For \( x = 4 \):
\[
h(4) = 45.3 + 9.3 \times 4 = 45.3 + 37.2 = 82.5
\]

Now we can complete the table:

\[
\begin{array}{|c|c|}
\hline
x & h(x) \\
\hline
0 & 45.3 \\
1 & 54.6 \\
2 & 63.9 \\
3 & 73.2 \\
4 & 82.5 \\
\hline
\end{array}
\]