Question 210183
Let's assume that there is a linear relationship. So this means that there's a line that goes through every point listed above.


To find the equation of the line, we only need two points. So let's use (0,3) and (2,4)





First let's find the slope of the line through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(2,4\right)]



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(0,3\right)]. So this means that {{{x[1]=0}}} and {{{y[1]=3}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(2,4\right)].  So this means that {{{x[2]=2}}} and {{{y[2]=4}}}.



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(4-3)/(2-0)}}} Plug in {{{y[2]=4}}}, {{{y[1]=3}}}, {{{x[2]=2}}}, and {{{x[1]=0}}}



{{{m=(1)/(2-0)}}} Subtract {{{3}}} from {{{4}}} to get {{{1}}}



{{{m=(1)/(2)}}} Subtract {{{0}}} from {{{2}}} to get {{{2}}}



So the slope of the line that goes through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(2,4\right)] is {{{m=1/2}}}



Now let's use the point slope formula:



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y-3=(1/2)(x-0)}}} Plug in {{{m=1/2}}}, {{{x[1]=0}}}, and {{{y[1]=3}}}



{{{y-3=(1/2)x-(1/2)(0)}}} Distribute



{{{y-3=(1/2)x+0}}} Multiply



{{{y=(1/2)x+0+3}}} Add 3 to both sides. 



{{{y=(1/2)x+3}}} Combine like terms. 



So the equation that goes through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(2,4\right)] is {{{y=(1/2)x+3}}}



This line also goes through the other points listed above.



Note: To make a function, simply replace 'y' with 'f(x)' to get {{{f(x)=(1/2)x+3}}}