A data set consists of the following data points:

1 answer:

4 0

(3,5),
(5,8),
(6,13)
------
(14,26)/3

then use the point slope form of a line to find equation of line

 $y-y_1=m(x-x_1)
\\y - \frac{26}{3} = 2.5(x- \frac{14}{3})
\\y - \frac{26}{3} = \frac{5}{2} (x- \frac{14}{3})
\\y - \frac{26}{3} = \frac{5}{2} x- \frac{5}{2}\times\frac{14}{3}
\\y - \frac{26}{3} = \frac{5}{2} x- \frac{35}{3}
\\6y-52=15x-70
\\15x-6y-18=0$

Find y-intercept for x = 0.

 $15x-6y-18=0
\\15\times0-6y-18=0
\\-6y=18
\\y=-3$

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Find the slope of the line graphed below.

Finger [1]

to get the equation of any straight line, we simply need two points off of it, let's use the ones provided in the picture below.

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}} \implies \cfrac{-7}{2}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{-\cfrac{7}{2}}(x-\stackrel{x_1}{1}) \\\\\\ y-4=-\cfrac{7}{2}x+\cfrac{7}{2}\implies y=-\cfrac{7}{2}x+\cfrac{7}{2}+4\implies y=-\cfrac{7}{2}x+\cfrac{15}{2}$