Question 112935
Since the x-intercept is -3 and the y intercept is 6, the line goes through the points (-3,0) and (0,6)



First lets find the slope through the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{6}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{-3}}},{{{0}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{0}}},{{{6}}}))


{{{m=(6-0)/(0--3)}}} Plug in {{{y[2]=6}}},{{{y[1]=0}}},{{{x[2]=0}}},{{{x[1]=-3}}}  (these are the coordinates of given points)


{{{m= 6/3}}} Subtract the terms in the numerator {{{6-0}}} to get {{{6}}}.  Subtract the terms in the denominator {{{0--3}}} to get {{{3}}}

  


{{{m=2}}} Reduce

  

So the slope is

{{{m=2}}}


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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-0=(2)(x--3)}}} Plug in {{{m=2}}}, {{{x[1]=-3}}}, and {{{y[1]=0}}} (these values are given)



{{{y-0=(2)(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



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


{{{y-0=2x+6}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}


{{{y=2x+6+0}}} Add {{{0}}} to  both sides to isolate y


{{{y=2x+6}}} Combine like terms {{{6}}} and {{{0}}} to get {{{6}}} 

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Answer:



So the equation of the line which goes through the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{6}}})  is:{{{y=2x+6}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=2}}} and the y-intercept is {{{b=6}}}


Notice if we graph the equation {{{y=2x+6}}} and plot the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{6}}}),  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, -10.5, 7.5, -6, 12,
graph(500, 500, -10.5, 7.5, -6, 12,(2)x+6),
circle(-3,0,0.12),
circle(-3,0,0.12+0.03),
circle(0,6,0.12),
circle(0,6,0.12+0.03)
) }}} Graph of {{{y=2x+6}}} through the points ({{{-3}}},{{{0}}}) and ({{{0}}},{{{6}}})


Notice how the two points lie on the line. This graphically verifies our answer.