The points in the table lie on a line. Find the slope of the line.

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-1 & -4 \\
\hline
0 & -1 \\
\hline
1 & 2 \\
\hline
2 & 5 \\
\hline
\end{tabular}

The slope is [tex]$\boxed{3}$[/tex].



Answer :

To find the slope of the line that passes through the given points [tex]\((-1, -4)\)[/tex], [tex]\( (0, -1)\)[/tex], [tex]\( (1, 2)\)[/tex], and [tex]\( (2, 5)\)[/tex], we can use the formula for the slope between any two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] on a line:

[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

We will use the first two points from the table to calculate the slope. The points are [tex]\((-1, -4)\)[/tex] and [tex]\( (0, -1)\)[/tex].

1. Identify the coordinates of the first point: [tex]\((-1, -4)\)[/tex]
[tex]\[ x_1 = -1, \quad y_1 = -4 \][/tex]

2. Identify the coordinates of the second point: [tex]\( (0, -1)\)[/tex]
[tex]\[ x_2 = 0, \quad y_2 = -1 \][/tex]

3. Substitute these values into the slope formula:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-4)}{0 - (-1)} \][/tex]

4. Simplify the expression:
[tex]\[ \text{slope} = \frac{-1 + 4}{0 + 1} = \frac{3}{1} = 3.0 \][/tex]

Therefore, the slope of the line passing through the given points is [tex]\( \boxed{3.0} \)[/tex].