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Which linear function has the same slope as the one represented by the table below?

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
$-\frac{1}{2}$ & $\frac{1}{5}$ \\
\hline
$-\frac{1}{5}$ & $\frac{7}{50}$ \\
\hline
$\frac{1}{5}$ & $\frac{3}{50}$ \\
\hline
$\frac{1}{2}$ & 0 \\
\hline
\end{tabular}
\][/tex]

Sagot :

To find the linear function that has the same slope as the one represented by the table, we need to calculate the slope using two of the given points. The points from the table are:
[tex]\[ (-\frac{1}{2}, \frac{1}{5}), \quad (-\frac{1}{5}, \frac{7}{50}), \quad (\frac{1}{5}, \frac{3}{50}), \quad (\frac{1}{2}, 0) \][/tex]

Let's use the first two points [tex]\((- \frac{1}{2}, \frac{1}{5})\)[/tex] and [tex]\((- \frac{1}{5}, \frac{7}{50})\)[/tex] to find the slope ([tex]\(m\)[/tex]) of the line.

The formula for the slope [tex]\(m\)[/tex] of a line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

Plugging in the coordinates of the first two points:
[tex]\[ x_1 = -\frac{1}{2}, \quad y_1 = \frac{1}{5} \\ x_2 = -\frac{1}{5}, \quad y_2 = \frac{7}{50} \][/tex]

Now, let's calculate the slope:
[tex]\[ m = \frac{\frac{7}{50} - \frac{1}{5}}{-\frac{1}{5} - (-\frac{1}{2})} \][/tex]

First, we need to have the same denominator in the numerator of our slope fraction:
[tex]\[ \frac{1}{5} = \frac{10}{50} \][/tex]
So,
[tex]\[ m = \frac{\frac{7}{50} - \frac{10}{50}}{-\frac{1}{5} - (-\frac{1}{2})} = \frac{\frac{7 - 10}{50}}{-\frac{1}{5} - (-\frac{1}{2})} = \frac{\frac{-3}{50}}{-\frac{1}{5} + \frac{1}{2}} \][/tex]

Next, let's calculate the denominator:
[tex]\[ -\frac{1}{5} + \frac{1}{2} \][/tex]

To add these two fractions, we need a common denominator, which is 10:
[tex]\[ -\frac{1}{5} = -\frac{2}{10}, \quad \frac{1}{2} = \frac{5}{10} \][/tex]

So,
[tex]\[ -\frac{1}{5} + \frac{1}{2} = -\frac{2}{10} + \frac{5}{10} = \frac{3}{10} \][/tex]

Now we can finalize our slope calculation:
[tex]\[ m = \frac{\frac{-3}{50}}{\frac{3}{10}} = \frac{-3}{50} \cdot \frac{10}{3} = -\frac{30}{150} = -0.2 \][/tex]

So, the slope [tex]\(m\)[/tex] of the line is [tex]\(-0.2\)[/tex].

Thus, the linear function that has the same slope as the one represented by the table has a slope [tex]\(m = -0.2\)[/tex]. The linear equation of the form [tex]\(y = mx + b\)[/tex] can have various [tex]\(b\)[/tex] values, depending on the y-intercept, but in our case, the slope [tex]\(m\)[/tex] is always [tex]\(-0.2\)[/tex].