Question 93107
Since it is a "linear" function, they are looking for an equation in the form of {{{y=mx+b}}} which forms a line. So we have the points (3,4) and (0,9).  Lets find the equation through those points



First lets find the slope through the points ({{{3}}},{{{4}}}) and ({{{0}}},{{{9}}})


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


{{{m=(9-4)/(0-3)}}} Plug in {{{y[2]=9}}},{{{y[1]=4}}},{{{x[2]=0}}},{{{x[1]=3}}}  (these are the coordinates of given points)


{{{m= 5/-3}}} Subtract the terms in the numerator {{{9-4}}} to get {{{5}}}.  Subtract the terms in the denominator {{{0-3}}} to get {{{-3}}}

  


{{{m=-5/3}}} Reduce

  

So the slope is

{{{m=-5/3}}}


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Now let's use the point-slope formula to find the equation of the line:




------Point-Slope Formula------
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(\textrm{x_{1},y_{1}}\right)] is one of the given points


So lets use the Point-Slope Formula to find the equation of the line


{{{y-4=(-5/3)(x-3)}}} Plug in {{{m=-5/3}}}, {{{x[1]=3}}}, and {{{y[1]=4}}} (these values are given)



{{{y-4=(-5/3)x+(-5/3)(-3)}}} Distribute {{{-5/3}}}


{{{y-4=(-5/3)x+5}}} Multiply {{{-5/3}}} and {{{-3}}} to get {{{15/3}}}. Now reduce {{{15/3}}} to get {{{5}}}


{{{y=(-5/3)x+5+4}}} Add {{{4}}} to  both sides to isolate y


{{{y=(-5/3)x+9}}} Combine like terms {{{5}}} and {{{4}}} to get {{{9}}} 

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Answer:



So the equation of the line which goes through the points ({{{3}}},{{{4}}}) and ({{{0}}},{{{9}}})  is:{{{y=(-5/3)x+9}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=-5/3}}} and the y-intercept is {{{b=9}}}


Notice if we graph the equation {{{y=(-5/3)x+9}}} and plot the points ({{{3}}},{{{4}}}) and ({{{0}}},{{{9}}}),  we get this: (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{drawing(500, 500, -7.5, 10.5, -2.5, 15.5,
graph(500, 500, -7.5, 10.5, -2.5, 15.5,(-5/3)x+9),
circle(3,4,0.12),
circle(3,4,0.12+0.03),
circle(0,9,0.12),
circle(0,9,0.12+0.03)
) }}} Graph of {{{y=(-5/3)x+9}}} through the points ({{{3}}},{{{4}}}) and ({{{0}}},{{{9}}})



Since the equation is {{{y=(-5/3)x+9}}} the function is {{{f(x)=(-5/3)x+9}}} (just replace y with f(x))