Sure! Let's work through this step by step.
### Part (a)
We have a table representing points of a linear function:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
0 & $\square$ \\
\hline
20 & 16 \\
\hline
40 & 28 \\
\hline
\end{tabular}
\][/tex]
First, we need to find the value of [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex]. This is essentially finding the y-intercept [tex]\( b \)[/tex] of the line.
We were given the points [tex]\((20, 16)\)[/tex] and [tex]\((40, 28)\)[/tex].
1. Calculate the slope [tex]\( m \)[/tex]:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{28 - 16}{40 - 20} = \frac{12}{20} = 0.6
\][/tex]
2. Find the y-intercept [tex]\( b \)[/tex]:
Use the point [tex]\((20, 16)\)[/tex] and the slope [tex]\( m \)[/tex] to find [tex]\( b \)[/tex]:
[tex]\[
y = mx + b \implies 16 = 0.6 \cdot 20 + b \implies 16 = 12 + b \implies b = 16 - 12 \implies b = 4
\][/tex]
Given [tex]\( x = 0 \)[/tex], the value of [tex]\( y \)[/tex] is:
[tex]\[
y = b = 4
\][/tex]
So, the table now looks like this:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
0 & 4 \\
\hline
20 & 16 \\
\hline
40 & 28 \\
\hline
\end{tabular}
\][/tex]
### Part (b)
Next, we need to report the equation of the line in slope-intercept form, [tex]\( y = mx + b \)[/tex].
We have already calculated the slope [tex]\( m \)[/tex] to be [tex]\( 0.6 \)[/tex] and the y-intercept [tex]\( b \)[/tex] to be [tex]\( 4 \)[/tex].
Thus, the equation of the line is:
[tex]\[
y = 0.6x + 4
\][/tex]
Therefore, our final answers are:
a. The value in the first row for [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex] is 4.
b. The equation of the line in slope-intercept form is [tex]\( y = 0.6x + 4 \)[/tex].
