Question

Complete the slope-intercept form of the linear equation that represents the relationship in the table.

x y
−1 −10
3 14
CLEAR CHECK
y=
x + (
)

Answer 1

to get the equation of any straight line, we simply need two points off of it, let's use those two points from the table
`\begin{array}c \cline{1-2} x&y\\ \cline{1-2} &\\ -1&-10\\ 3&14\\ \cline{1-2} \end{array}`

`(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-10})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{14}-\stackrel{y1}{(-10)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-1)}}} \implies \cfrac{14 +10}{3 +1} \implies \cfrac{ 24 }{ 4 }\implies 6`

`\begin{array}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}{(-10)}=\stackrel{m}{6}(x-\stackrel{x_1}{(-1)}) \implies y +10= 6 (x +1) \\\\\\ y+10=6x+6\implies {\Large \begin{array}{llll} y=6x-4 \end{array}}`