To find the equation of the linear function that best fits the given data using linear regression, we will calculate the slope ($m$) and the y-intercept ($b$) of the line in the form $y = mx + b$.

Given data points: $x: 1, 2, 3, 4, 5, 6$ $y: 86, 108, 126, 160, 181, 193$

Step 1: Calculate the means of $x$ and $y$

$\bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5$ $\bar{y} = \frac{86 + 108 + 126 + 160 + 181 + 193}{6} = \frac{854}{6} \approx 142.33$

Step 2: Calculate the slope ($m$)

The formula for the slope $m$ is: $m = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}$

Let's compute the sums needed: $\sum{(x_i - \bar{x})(y_i - \bar{y})} = (1-3.5)(86-142.33) + (2-3.5)(108-142.33) + \dots + (6-3.5)(193-142.33)$ $= (-2.5)(-56.33) + (-1.5)(-34.33) + (-0.5)(-16.33) + (0.5)(17.67) + (1.5)(38.67) + (2.5)(50.67)$ $= 140.825 + 51.495 + 8.165 + 8.835 + 58.005 + 126.675 = 394.0$

$\sum{(x_i - \bar{x})^2} = (1-3.5)^2 + (2-3.5)^2 + \dots + (6-3.5)^2$ $= (-2.5)^2 + (-1.5)^2 + (-0.5)^2 + (0.5)^2 + (1.5)^2 + (2.5)^2$ $= 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5$

$m = \frac{394.0}{17.5} \approx 22.51$

Step 3: Calculate the y-intercept ($b$)

The y-intercept $b$ is calculated using the formula: $b = \bar{y} - m\bar{x}$ $b = 142.33 - 22.51 \times 3.5 \approx 142.33 - 78.785 \approx 63.54$

Step 4: Write the final equation

The linear equation that best fits the data is: $y = 22.51x + 63.54$

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. How would the linear regression change if an outlier was added to the data set?
2. How can you determine the strength of the fit in a linear regression?
3. What are some real-world applications of linear regression?
4. How does linear regression differ from multiple regression?
5. Can linear regression be used to predict values outside the given range of data?

Tip: In linear regression, it's important to check if the relationship between the variables is actually linear, as using a linear model for nonlinear data can lead to inaccurate predictions.