Question

Find the equation of the linear function represented by the table below in slope-

intercept form.
x
o
1
2
3
4
у
4
10
16
22
28

Answer 1

to get the equation of any straight line we simply need two points off of it, hmm let's get two from this table

`\begin{array}cc \cline{1-2} x&y\\ \cline{1-2} 0&4\\ 1&10\\ 2&16\\ 3&22\\ 4&28\\ \cline{1-2} \end{array} \begin{array}{llll} \\ \leftarrow \textit{let's use this point}\\\\ \leftarrow \textit{and this point} \end{array}`

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{22}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{22}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{12}{2}\implies 6 \\\\\\ \begin{array}c \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}{10}=\stackrel{m}{6}(x-\stackrel{x_1}{1}) \\\\\\ y-10=6x-6\implies y=6x+4`