To create our linear equation, we are going to use the tow points. Our first point will be the first point in our data graph, so (8, 525), and our second point will be the last point in our data graph, so (22, 2395). The [tex]x-axis[/tex] will represent the time, and the [tex]y-axis[/tex] will represent the calories.
The first thing we are going to do is find the slope of our linear equation. To do that we'll use the slope formula: [tex]m= \frac{y_{2}-y_{1} }{x_{2}-x_{1}} [/tex]. For our two points we can infer that [tex]x_{1}=8[/tex], [tex]y_{1}=525[/tex], [tex]x_{2}=22[/tex], and [tex]y_{2}=2395[/tex]. So lets replace those values in our slope formula to find [tex]m[/tex]:
[tex]m= \frac{2395-525}{22-8} [/tex]
[tex]m= \frac{1870}{14} [/tex]
[tex]m= \frac{935}{7} [/tex]
Now that we have the slope, we are going to use the point slope formula, [tex]y-y_{1}=m(x-x_{1})[/tex], to complete our equation:
[tex]y-525= \frac{935}{7} (x-8)[/tex]
[tex]y-525= \frac{935}{7} x- \frac{7480}{7} [/tex]
[tex]y= \frac{935}{7} x- \frac{7480}{7} +525[/tex]
[tex]y= \frac{935}{7} x- \frac{3805}{7} [/tex]
We can conclude that the linear equation that best fits our data is [tex]y= \frac{935}{7} x- \frac{3805}{7} [/tex].