Answer:
[tex]y=\displaystyle\frac{1}{2} x-3[/tex]
Step-by-step explanation:
Hi there!
Linear equations are typically organized in slope-intercept form: [tex]y=mx+b[/tex] where m is the slope and b is the y-intercept (the value of y when x=0)
1) Determine the slope (m)
[tex]m=\displaystyle\frac{y_2-y_1}{x_2-x_1}[/tex] where two points on the line are [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]
Plug in the given points (4,-1) and (0,3)
[tex]m=\displaystyle\frac{-1-(-3)}{4-0}\\\\m=\displaystyle\frac{2}{4}\\\\m=\displaystyle\frac{1}{2}[/tex]
Therefore, the slope of the line is [tex]\displaystyle\frac{1}{2}[/tex]. Plug this into [tex]y=mx+b[/tex] :
[tex]y=\displaystyle\frac{1}{2} x+b[/tex]
2) Determine the y-intercept (b)
[tex]y=\displaystyle\frac{1}{2} x+b[/tex]
The y-intercept occurs when x=0. Because we're given that (0,-3), the y-intercept is therefore -3.
[tex]y=\displaystyle\frac{1}{2} x-3[/tex]
I hope this helps!
