Question

raquel throws darts at a coordinate grid centered at the origin. her goal is to create a line of darts. her darts actually hit the coordinate grid at (-5, 0), (1, -3), (4, 5), (-8, -6), (0, 2), and (9, 6). which equation best approximates the line of best fit of the darts?
o y = 0.6x + 0.6
o y = 0.1x + 0.8
o y = 0.8x + 0.1
o y = 0.5x + 0.6

Explanation:

Step1: Recall the form of a linear - regression line

The equation of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. We can use a statistical method (or graphing calculator) to find the line of best fit. Another way is to use the fact that we can check the average behavior of the points. We can calculate the means of the $x$ - values and $y$ - values of the given points.
Let the points be $(x_1,y_1)=(-5,0),(x_2,y_2)=(1, - 3),(x_3,y_3)=(4,5),(x_4,y_4)=(-8,-6),(x_5,y_5)=(0,2),(x_6,y_6)=(9,6)$.
The mean of the $x$ - values $\bar{x}=\frac{-5 + 1+4-8 + 0+9}{6}=\frac{1}{6}\approx0.17$.
The mean of the $y$ - values $\bar{y}=\frac{0-3 + 5-6 + 2+6}{6}=\frac{4}{6}\approx0.67$.
The slope $m$ of the line of best fit can be estimated by looking at the general trend of how $y$ changes with $x$.
We can also use the fact that for a line $y=mx + b$, when $x = 0$, $y=b$. We can check which equation gives a reasonable approximation for the points.
Let's take two points, say $(0,2)$ and $(4,5)$. The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{5 - 2}{4-0}=\frac{3}{4}=0.75$.
A more accurate way is to use a calculator with linear - regression capabilities. But by estimating:
We know that when $x = 0$, we can get an idea of the $y$ - intercept. Among the given equations, when we consider the general trend of the points and the fact that the line should pass through the "center" of the cloud of points.
We can substitute some of the points into the equations to check.
For the point $(0,2)$, substituting into $y = 0.6x+0.6$, we get $y = 0.6$ (not a good fit).
For the point $(0,2)$, substituting into $y = 0.1x + 0.8$, we get $y=0.8$ (not a good fit).
For the point $(0,2)$, substituting into $y = 0.8x+0.1$, we get $y = 0.1$ (not a good fit).
For the point $(0,2)$, substituting into $y = 0.5x+0.6$, when $x = 0$, $y = 0.6$. When we consider the general trend of the other points and the fact that the slope of around $0.5$ seems to follow the general increase of $y$ with $x$ for the given points.

Step2: Check with more points

Let's take the point $(4,5)$.
For $y = 0.5x+0.6$, when $x = 4$, $y=0.5\times4 + 0.6=2 + 0.6=2.6$ (not a perfect fit but closer compared to other equations).
Let's take the point $(1,-3)$. For $y = 0.5x+0.6$, when $x = 1$, $y=0.5\times1+0.6 = 1.1$ (not a great fit but still, among the options, this equation seems to be the best approximation).

Answer:

$y = 0.6x+0.6$